assignment makes integer from pointer without a cast null

Converting Pointers to Integers: Avoiding Cast Errors & Mastering the Process

In this guide, we will cover how to convert pointers to integers and vice versa without running into any cast errors. We will also walk you through the step-by-step process of mastering pointer and integer conversions in C/C++.

Why Convert Pointers to Integers
Understanding uintptr_t
Step-by-Step Guide
Converting Pointers to Integers
Converting Integers to Pointers

Why Convert Pointers to Integers? {#why-convert-pointers-to-integers}

There are several use cases where you might need to convert pointers to integers and vice versa. Some common reasons include:

Manipulating memory addresses for low-level programming.
Serializing and deserializing data.
Storing pointers in a generic data structure.
Debugging and logging purposes.

However, when converting pointers to integers, it is crucial to avoid any errors that may arise from incorrect casting.

Understanding uintptr_t {#understanding-uintptr_t}

To safely convert pointers to integers, it is essential to use the uintptr_t data type. This is an unsigned integer type that is large enough to store the value of a pointer. It is available in the <stdint.h> header in C and the <cstdint> header in C++.

Using uintptr_t , you can safely cast a pointer to an integer and back to a pointer without losing any information. This ensures that the process is safe, fast, and efficient.

Step-by-Step Guide {#step-by-step-guide}

Converting pointers to integers {#converting-pointers-to-integers}.

To convert a pointer to an integer, follow these steps:

Include the <stdint.h> header (C) or the <cstdint> header (C++) in your program.
Cast your pointer to uintptr_t .

Converting Integers to Pointers {#converting-integers-to-pointers}

To convert an integer to a pointer, follow these steps:

Cast your integer to the required pointer type using a double cast.

FAQs {#faqs}

Why can't i just use a regular int or unsigned int to store pointers {#regular-int}.

While it may work on some platforms where the size of an int is equal to the size of a pointer, it is not guaranteed to be portable across different systems. Using uintptr_t ensures your code remains portable and safe.

Are there performance implications when using uintptr_t ? {#performance}

The performance impact of using uintptr_t is minimal. Most modern compilers can optimize the casting operations, resulting in little to no overhead.

When should I use intptr_t instead of uintptr_t ? {#intptr_t}

intptr_t is a signed integer type that can hold a pointer value. It is useful when you need to perform arithmetic operations on pointers that may result in negative values. However, in most cases, uintptr_t is recommended.

Is it safe to perform arithmetic operations on integers representing pointers? {#pointer-arithmetic}

Performing arithmetic operations on integers representing pointers can lead to undefined behavior if the resulting integer doesn't correspond to a valid memory address. It is generally safer to perform arithmetic operations on pointers directly.

How do I avoid losing information when casting pointers to integers? {#avoid-losing-information}

By using uintptr_t , you ensure that the integer is large enough to store the value of a pointer without losing any information. Make sure always to use uintptr_t when converting pointers to integers.

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Posted on Oct 4, 2023 • Originally published at coderlegion.com

Assignment makes integer from pointer without a cast in c

Programming can be both rewarding and challenging. You work hard on your code, and just when it seems to be functioning perfectly, an error message pops up on your screen, leaving you frustrated and clueless about what went wrong. One common error that programmers encounter is the "Assignment makes integer from pointer without a cast" error in C.

This error occurs when you try to assign a value from a pointer variable to an integer variable without properly casting it. To fix this error, you need to make sure that you cast the pointer value to the appropriate data type before assigning it to an integer variable. In this article, we will dive deeper into the causes of this error and provide you with solutions to overcome it.

What makes this error occur?

I will present some cases that triggers that error to occur, and they are all have the same concept, so if you understanded why the failure happens, then you will figure out how to solve all the cases easily.

Case 1: Assignment of a pointer to an integer variable

In this simple code we have three variables, an integer pointer "ptr", and two integers "n1" and "n2". We assign 2 to "n1", so far so good, then we assign the address of "n2" to "ptr" which is the suitable storing data type for a pointer, so no problems untill now, till we get to this line "n2 = ptr" when we try to assign "ptr" which is a memory address to "n2" that needs to store an integer data type because it's not a pointer.

Case 2: Returning a Pointer from a Function that Should Return an Integer

As you can see, it's another situation but it's the same idea which causes the compilation error. We are trying here to assign the return of getinteger function which is a pointer that conatains a memory address to the result variable that is an int type which needs an integer

Case 3: Misusing Array Names as Pointers

As we might already know, that the identifier (name) of the array is actually a pointer to the array first element memory address, so it's a pointer after all, and assigning a pointer type to int type causes the same compilation error.

The solutions

The key to avoiding the error is understanding that pointers and integers are different types of variables in C. Pointers hold memory addresses, while integers hold numeric values. We can use either casting, dereferencing the pointer or just redesign another solution for the problem we are working on that allows the two types to be the same. It all depending on the situation.

Let's try to solve the above cases:

Case 1: Solution: Deferencing the pointer

We need in this case to asssign an int type to "n2" not a pointer or memory address, so how do we get the value of the variable that the pointer "ptr" pointing to? We get it by deferencing the pointer, so the code after the fix will be like the following:

Case 2: Solution: Choosing the right data type

In this case we have two options, either we change the getinteger returning type to int or change the result variable type to a pointer. I will go with the latter option, because there are a lot of functions in the C standard library that returning a pointer, so what we can control is our variable that takes the function return. So the code after the fix will be like the following:

We here changed the result variable from normal int to an int pointer by adding "*".

Case 3: Solution: Using the array subscript operator

In this case we can get the value of any number in the array by using the subscript opeartor ([]) on the array with the index number like: myarray[1] for the second element which is 2. If we still remember that the array identifier is a pointer to the array first memory, then we can also get the value of the array first element by deferencing the array identifier like: *myarray which will get us 1.

But let's solve the case by using the subscript opeartor which is the more obvious way. So the code will be like the following:

Now the number 1 is assigned to myint without any compilation erros.

The conclusion

In conclusion, the error "assignment to ‘int’ from ‘int *’ makes integer from pointer without a cast" arises in C programming when there is an attempt to assign a memory address (held by a pointer) directly to an integer variable. This is a type mismatch as pointers and integers are fundamentally different types of variables in C.

To avoid or correct this error, programmers need to ensure they are handling pointers and integers appropriately. If the intent is to assign the value pointed by a pointer to an integer, dereferencing should be used. If a function is meant to return an integer, it should not return a pointer. When dealing with arrays, remember that the array name behaves like a pointer to the first element, not an individual element of the array.

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warning:assignment makes integer from po

warning:assignment makes integer from pointer without a cast.

What is Assignment Makes Pointer From Integer Without A Cast and How Can It Be Avoided?

Understanding the pointer-integer conundrum in programming.

In the realm of programming, particularly in languages like C and C++, pointers play a crucial role in managing memory and optimizing the performance of applications. However, they can also be a source of confusion and errors, especially for those new to these languages. One common error that programmers encounter is the “assignment makes pointer from integer without a cast” warning or error. This message can be perplexing, but understanding its meaning is essential for writing clean, efficient, and error-free code.

Decoding the Warning: Assignment Makes Pointer from Integer Without a Cast

The warning “assignment makes pointer from integer without a cast” is a compiler message that indicates a potential issue in your code. It occurs when an integer value is assigned to a pointer variable without explicitly casting the integer to a pointer type. This can happen for various reasons, such as mistakenly using an integer as a memory address or overlooking the need for a type conversion.

Why Does This Warning Matter?

Ignoring this warning can lead to undefined behavior in your program. Since pointers and integers are fundamentally different types, treating an integer as a memory address can cause your program to access memory locations it shouldn’t, leading to crashes, data corruption, or security vulnerabilities.

Examples of Pointer-Integer Assignment Issues

To better understand the warning, let’s look at some examples where this issue might arise:

In the example above, the variable num is an integer, and ptr is a pointer to an integer. The assignment of num to ptr without a cast triggers the warning.

Root Causes of the Pointer-Integer Assignment Warning

Several scenarios can lead to this warning. Understanding these scenarios can help you avoid the issue in your code.

Accidental assignment of an integer to a pointer variable.
Using an integer as a function return value where a pointer is expected.
Incorrectly using an integer constant as a null pointer.
Forgetting to allocate memory before assigning it to a pointer.

Strategies to Avoid Pointer-Integer Assignment Issues

To prevent the “assignment makes pointer from integer without a cast” warning, you can employ several strategies:

Always use explicit casting when converting an integer to a pointer.
Ensure that functions returning pointers do not accidentally return integers.
Use the NULL macro or nullptr (in C++) for null pointers instead of integer constants.
Properly allocate memory using functions like malloc or new before assigning it to pointers.

Using Explicit Casting

Explicit casting involves converting one data type to another by specifying the target type. In the context of pointers and integers, you can use a cast to tell the compiler that you intentionally want to treat an integer as a pointer.

Proper Function Return Types

When writing functions that are supposed to return pointers, ensure that the return type is correctly declared as a pointer type, and that the return statements within the function adhere to this type.

Using NULL or nullptr

Instead of using integer constants like 0 for null pointers, use the NULL macro in C or nullptr in C++ to avoid confusion and potential warnings.

Memory Allocation

Before assigning a pointer, ensure that it points to a valid memory location by using memory allocation functions.

Advanced Considerations and Best Practices

Beyond the basic strategies, there are advanced considerations and best practices that can help you write robust pointer-related code:

Use static analysis tools to catch potential pointer-related issues before runtime.
Adopt coding standards that discourage unsafe practices with pointers and integers.
Understand the underlying architecture and how pointers are represented and used.
Regularly review and refactor code to improve clarity and safety regarding pointer usage.

FAQ Section

What is a pointer in programming.

A pointer is a variable that stores the memory address of another variable. Pointers are used for various purposes, such as dynamic memory allocation, implementing data structures like linked lists, and optimizing program performance.

Why is casting necessary when assigning an integer to a pointer?

Casting is necessary because pointers and integers are different data types with different interpretations. A cast explicitly informs the compiler that the programmer intends to treat the integer as a pointer, which can prevent unintended behavior.

Can ignoring the pointer-integer assignment warning lead to security issues?

Yes, ignoring this warning can lead to security issues such as buffer overflows, which can be exploited by attackers to execute arbitrary code or cause a program to crash.

Is it always safe to cast an integer to a pointer?

No, it is not always safe. The integer should represent a valid memory address that the program is allowed to access. Arbitrary casting without ensuring this can lead to undefined behavior and program crashes.

What is the difference between NULL and nullptr?

NULL is a macro that represents a null pointer in C and is typically defined as 0 or ((void *)0). nullptr is a keyword introduced in C++11 that represents a null pointer and is type-safe, meaning it cannot be confused with integer types.

The “assignment makes pointer from integer without a cast” warning is a signal from the compiler that there’s a potential issue in your code. By understanding what causes this warning and how to avoid it, you can write safer and more reliable programs. Always be mindful of the types you’re working with, use explicit casting when necessary, and follow best practices for pointer usage. With these strategies in place, you’ll be well-equipped to handle pointers and integers in your coding endeavors.

For further reading and a deeper understanding of pointers, integer casting, and related topics, consider exploring the following resources:

C Programming Language (2nd Edition) by Brian W. Kernighan and Dennis M. Ritchie
Effective Modern C++ by Scott Meyers
C++ Primer (5th Edition) by Stanley B. Lippman, Josée Lajoie, and Barbara E. Moo
ISO/IEC 9899:201x – International Standard for Programming Language C
ISO/IEC 14882:2017 – International Standard for Programming Language C++

These references provide a solid foundation for understanding the intricacies of pointers and type casting, as well as best practices for writing secure and efficient code in C and C++.

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[SOLVED] C - assigment makes integer from pointer without a cast warning

Thread: [solved] c - assigment makes integer from pointer without a cast warning, thread tools.

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I know it's a common error, and I've tried googling it, looked at a bunch of answers, but I still don't really get what to do in this situation.... Here's the relevant code: Code: #include <stdio.h> #include <stdlib.h> #include <string.h> int main( int argc, char *argv[] ) { char path[51]; const char* home = getenv( "HOME" ); strcpy( path, argv[1] ); path[1] = home; return 0; } -- there is more code in the blank lines, but the issue's not there (I'm fairly sure), so didn't see the point in writing out 100 odd lines of code. I've tried some stuff like trying to make a pointer to path[1], and make that = home, but haven't managed to make that work (although maybe that's just me doing it wrong as opposed to wrong idea?) Thanks in advance for any help

Re: C - assigment makes integer from pointer without a cast warning

path[1] is the second element of a char array, so it's a char. home is a char *, i.e. a pointer to a char. You get the warning because you try to assign a char* to a char. Note also the potential to overflow your buffer if the content of the argv[1] argument is very long. It's usually better to use strncpy.

Last edited by r-senior; March 10th, 2013 at 03:03 PM . Reason: argv[1] would overflow, not HOME. Corrected to avoid confusion.

Please create new threads for new questions. Please wrap code in code tags using the '#' button or enter it in your post like this: [code]...[/code].

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You can try something like this: Code: #include <stdio.h> #include <stdlib.h> #include <string.h> int main( int argc, char *argv[] ) { char *path; const char *home = getenv("HOME"); path = malloc(strlen(home) + 1); if (!path) { printf("Error\n"); return 0; } strcpy(path, home); printf("path = %s\n", path); // if you want to have argv[1] concatenated with path if (argc >= 2) { path = malloc(strlen(home) + strlen(argv[1]) + 1); strcpy(path, argv[1]); strcat(path, home); printf("%s\n", path); } // if you want an array of strings, each containing path, argv[1]... char **array; int i; array = malloc(argc * sizeof(char*)); array[0] = malloc(strlen(home) + 1); strcpy(array[0], home); printf("array[0] = %s\n", array[0]); for (i = 1; i < argc; i++) { array[i] = malloc(strlen(argv[i]) + 1); strcpy(array[i], argv[i]); printf("array[%d] = %s\n", i, array[i]); } // now array[i] will hold path and all the argv strings return 0; } Just as above, your path[51] is a string while path[1] is only a character, so you can't use strcpy for that.

Last edited by Christmas; March 10th, 2013 at 09:51 PM .

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Originally Posted by Christmas You can try something like this: Code: #include <stdio.h> #include <stdlib.h> #include <string.h> int main( int argc, char *argv[] ) { char *path; const char *home = getenv("HOME"); path = malloc(strlen(home) + 1); if (!path) { printf("Error\n"); return 0; } strcpy(path, home); printf("path = %s\n", path); // if you want to have argv[1] concatenated with path if (argc >= 2) { path = malloc(strlen(home) + strlen(argv[1]) + 1); strcpy(path, argv[1]); strcat(path, home); printf("%s\n", path); } // if you want an array of strings, each containing path, argv[1]... char **array; int i; array = malloc(argc * sizeof(char*)); array[0] = malloc(strlen(home) + 1); strcpy(array[0], home); printf("array[0] = %s\n", array[0]); for (i = 1; i < argc; i++) { array[i] = malloc(strlen(argv[i]) + 1); strcpy(array[i], argv[i]); printf("array[%d] = %s\n", i, array[i]); } // now array[i] will hold path and all the argv strings return 0; } Just as above, your path[51] is a string while path[1] is only a character, so you can't use strcpy for that. Excellent point. I've basically fixed my problem by reading up on pointers again (haven't done any C for a little while, so forgot some stuff), and doing: Code: path[1] = *home; the code doesn't moan at me when I compile it, and it runs okay (for paths which aren't close to 51 at least), but after reading what you read, I just wrote a quick program and found out that getenv("HOME") is 10 characters long, not 1 like I seem to have assumed, so I'll modify my code to fix that.

Yes, getenv will return the path to your home dir, for example /home/user, but path[1] = *home will still assign the first character of home to path[1] (which would be '/').

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Assignment makes integer from pointer without a cast [-Werror]

I keep getting this error message in response to hashtable[i]:

assignment makes integer from pointer without a cast [-Werror]

Best Answer

If hashtable is an array of integers then hashtable[i] expects an integer and NULL is a pointer.

So you're trying to assign a pointer value to an integer variable(without a cast), this is usually just a warning but since you have -Werror all warnings turn into errors.

Just use 0 instead of NULL .

Sqlite – Improve INSERT-per-second performance of SQLite

Several tips:

Put inserts/updates in a transaction.
For older versions of SQLite - Consider a less paranoid journal mode ( pragma journal_mode ). There is NORMAL , and then there is OFF , which can significantly increase insert speed if you're not too worried about the database possibly getting corrupted if the OS crashes. If your application crashes the data should be fine. Note that in newer versions, the OFF/MEMORY settings are not safe for application level crashes.
Playing with page sizes makes a difference as well ( PRAGMA page_size ). Having larger page sizes can make reads and writes go a bit faster as larger pages are held in memory. Note that more memory will be used for your database.
If you have indices, consider calling CREATE INDEX after doing all your inserts. This is significantly faster than creating the index and then doing your inserts.
You have to be quite careful if you have concurrent access to SQLite, as the whole database is locked when writes are done, and although multiple readers are possible, writes will be locked out. This has been improved somewhat with the addition of a WAL in newer SQLite versions.
Take advantage of saving space...smaller databases go faster. For instance, if you have key value pairs, try making the key an INTEGER PRIMARY KEY if possible, which will replace the implied unique row number column in the table.
If you are using multiple threads, you can try using the shared page cache , which will allow loaded pages to be shared between threads, which can avoid expensive I/O calls.
Don't use !feof(file) !

I've also asked similar questions here and here .

ERROR: assignment makes integer from pointer without a cast

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Code: char *mid(char *tString, int x1, int x2){ char *rString; int i, l; rString= (char *) malloc(sizeof(char) * x2); l = x1-1; for (i= 0; i<x2; i++) { rString[i] = tString[l]; l++; } rString[x2]= NULL; return rString; } Mingw is giving me this error: C:\Users\Harshvardhan\Desktop\RPL\functions.h|30|w arning: assignment makes integer from pointer without a cast [enabled by default]| and for this line: Code: rString[x2]= NULL; And possible reason/cure for this?

rString is a pointer (presumably to an array of chars or a string). Therefore rString[x2] is a single char. But NULL is a pointer type, which is not compatible with char. You are probably looking for the null byte/character (yes, slightly confusing) to terminate your string. Try Code: rString[x2] = '\0';

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NULL is a null pointer constant. While it could just be the integer 0, it could also be (void*)0 or something else that ultimately is a null pointer constant. What you really want is a character constant, so just write: Code: rString[x2] = '\0'; By the way, as a single letter name, l is a horrible name as it can be confused with 1 on some fonts.

Originally Posted by Bjarne Stroustrup (2000-10-14) I get maybe two dozen requests for help with some sort of programming or design problem every day. Most have more sense than to send me hundreds of lines of code. If they do, I ask them to find the smallest example that exhibits the problem and send me that. Mostly, they then find the error themselves. "Finding the smallest program that demonstrates the error" is a powerful debugging tool. Look up a C++ Reference and learn How To Ask Questions The Smart Way

lol... And I always thought that NULL is exactly same as '\0'

Originally Posted by ehitam I always thought that NULL is exactly same as '\0' On some implementations, it is. Except for spelling, 0 is identical to '\0'. But you should not rely on that.

Originally Posted by laserlight By the way, as a single letter name, l is a horrible name as it can be confused with 1 on some fonts. I agree... I used it to signify length... And in code::blocks it is same as 1. I think I should change it with something better. And... Thanks laserlight and anduril for help...

Originally Posted by laserlight NULL is a null pointer constant. While it could just be the integer 0, it could also be (void*)0 or something else that ultimately is a null pointer constant. What you really want is a character constant, so just write: Code: rString[x2] = '\0'; By the way, as a single letter name, l is a horrible name as it can be confused with 1 on some fonts. Is this a past end of array? In other words, is the code below correct? Code: rString[x2-1] = '\0'; Really tired, so my thinking might be wrong. Tim S.

"...a computer is a stupid machine with the ability to do incredibly smart things, while computer programmers are smart people with the ability to do incredibly stupid things. They are,in short, a perfect match.." Bill Bryson

It looks more likely that the malloc should have allocated another char.

Originally Posted by stahta01 Is this a past end of array? No... x2-1 just make a character less... x2 is fine.

Originally Posted by laserlight It looks more likely that the malloc should have allocated another char. Do you mean rString?

Originally Posted by ehitam x2 is fine. The problem is that stahta01 is correct to say that it is a problem: you're writing to the array out of bounds. If you want to do this, your malloc should have a +1.

Originally Posted by laserlight The problem is that stahta01 is correct to say that it is a problem: you're writing to the array out of bounds. If you want to do this, your malloc should have a +1. Sorry... ... I'm not getting what you mean to say... When I use x2-1.... I get a character less...

Yes, you get a character less, but it's better than undefined behavior (what array out of bounds accesses result in). Arrays in C start at index 0 and go to index SIZE-1, or in your case x2-1. That is, your current code allocates an array of x2 characters, meaning you have rString[0]...rString[x2-1]. That's it, no more. rString[x2] is out of bounds. Tim's code corrects that in one possible way (by stopping at x2-1). Laserlight suggested another way, which allocates x2+1 characters, so rString[x2] is valid.

I mean that instead of this: Code: rString= (char *) malloc(sizeof(char) * x2); you should write: Code: rString = malloc(sizeof(*rString) * x2 + 1); Notice the +1 to account for the null character.

oh! Really didn't noticed that Thank you for help... I would replace x2 with x2+1

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How to create a shared library on Linux with GCC - December 30, 2011
Enum classes and nullptr in C++11 - November 27, 2011
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Rvalue References and Move Semantics in C++11 - November 13, 2011
C and C++ for Java Programmers - November 5, 2011
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