Hi,
I have an error warning: return makes integer from pointer without a cast. My code below
Code:#include<stdio.h> int fun ( int *p ) { *p = 20; return p; } int main () { int n; int a = 20; int *ptr =&a; n = fun (ptr); return 0; }
Can somebody help me?
Maybe this instead.
return *p;
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
Originally Posted by Salem
outputCode:#include<stdio.h> int fun ( int *p ) { *p = 20; return *p; } int main () { int n; int a = 20; int *ptr =&a; n = fun (ptr); printf(" a address = %p \n", &a); printf(" a content = %d \n", a); printf(" ptr = %p \n", ptr); printf(" ptr content = %d \n", *ptr); printf(" ptr address = %p \n", &ptr); printf(" n address content = %p \n ", &n); printf(" n content = %d \n", n); return 0; }
a address = 0061FF28
a content = 20
ptr = 0061FF28
ptr content = 20
ptr address = 0061FF24
n address content = 0061FF2C
n content = 20
I have rewritten
0061FF28 address a = 20
0061FF24 address ptr = 0061FF28
0061FF2C address n = 20
What the function fun return ?
20 or 0061FF28 ?
It's returning int, so it's 20.
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
but both 20 or 0061FF28 are integer
I am trying to make a function that takes the address of the variable and returns the address.
First example code how to get return address 0061FF28 from function ?
return p gives warning
Only mathematically. In this context, 20 is an int, whereas 0061FF28 is an address (i.e., the value of a pointer). The function is declared as returning an int, not a pointer.
Since you want to return a pointer, declare the function as returning a pointer, i.e., change the return type to int*
However, this also means that you should not store the return value of the function call in n since n is an int, not an int*
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How to verify that function return pointer
a address = 0061FF2CCode:#include<stdio.h> int * fun ( int *p ) { *p = 20; return p; } int main () { int n; int a = 20; int *ptr =&a; printf(" a address = %p \n", &a); printf(" a content = %d \n", a); printf(" ptr = %p \n", ptr); printf(" ptr content = %d \n", *ptr); printf(" ptr address = %p \n", &ptr); return 0; }
a content = 20
ptr = 0061FF2C
ptr content = 20
ptr address = 0061FF28
Check the return type of the function. Is it a pointer type? If yes, then the function returns a pointer.
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I don't have any idea how to do this
int * p, pointer type integer
Look at this example that you yourself wrote:
Here is the same function, but with the place in the syntax where the return type is clearly marked:Code:int fun ( int *p ) { *p = 20; return p; }
Code:LOOK_HERE_FOR_THE_RETURN_TYPE fun ( int *p ) { *p = 20; return p; }
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a address = 0061FF28Code:#include<stdio.h> int *fun ( int *p ) { *p = 20; return p; } int main () { int * n; int a = 20; int *ptr =&a; n = fun (ptr); printf(" a address = %p \n", &a); printf(" a content = %d \n", a); printf(" ptr = %p \n", ptr); printf(" ptr content = %d \n", *ptr); printf(" ptr address = %p \n", &ptr); printf(" n content = %p \n", n); return 0; }
a content = 20
ptr = 0061FF28
ptr content = 20
ptr address = 0061FF24
n content = 0061FF28
Does this make sense ?
Almost. To be consistent, the last printf should be:
Also, technically the argument corresponding to %p should be a void*, so you may see:Code:printf(" n = %p \n", n);
Code:printf(" n = %p \n", (void*)n);
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