assignment makes integer from pointer without a cast strcpy

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Error "'strcpy' makes pointer from integer without a cast"

The above code is just part of my full code. When I compiled my full program happened an error strcpy' makes pointer from integer without a cast which is at line

I have referred to the post at here makes pointer from integer without a cast with strcpy after I tried to use strncpy and follow the method showed in the post I still fail in compilation.Hope you can guide me with my problem.Thanks.

• compiler-errors

• you try to copy something on a int and not on a char * –  Alexis Commented Jul 23, 2013 at 6:54
• 1 Looks like you're trying to copy one int to another. Can't you just go: p3int = _PhaseShiftPin[i]->_PSPvalue; –  nemasu Commented Jul 23, 2013 at 6:57
• Please go through This link again and have a look on type of arguments. –  Dayal rai Commented Jul 23, 2013 at 6:58
• try char *p3int = NULL instead of int p3int = 0 ... –  someone Commented Jul 23, 2013 at 6:59

2 Answers 2

strcpy is inappropriate for this problem. It takes two arguments. The first is a pointer to char , and the second is a pointer to const char . It then copies the string pointed to by the second pointer into the memory pointed to by the first.

In your case, you are not copying a string, you are copying an int. Your last line should be:

However, the C99 standard reserves identifiers that begin with an underscore ('_') followed by an uppercase letter or another underscore for the implementation. Therefore you must not name your identifiers this way.

If you look at the definition of strcpy you can see

But your _PhaseShiftPin[i]->_PSPvalue is an integer and p3int also.

you can use memcpy

or simply as said in a comment

if you try to copy p3int into _PhaseShiftPin[i]->_PSPvalue

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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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