sizeof operation on char array

Question

Possible Duplicate:
How to find the sizeof(a pointer pointing to an array)
Sizeof an array in the C programming language?

#include<stdio.h>

void doit(char x[10]){
    printf("%d\n", sizeof(x));
}

void main(void){
    char x[10];
    printf("%d\n", sizeof(x));
    doit(x);
}

** I don't know why my question is removed first time. ** Two outputs here are different. Apparently the first one knows x is an array and second one only knows it a ptr. My question is why compiler knows that in the first case it's an array instead of a ptr?

http://stackoverflow.com/questions/492384/how-to-find-the-sizeofa-pointer-pointing-to-an-array — cnicutar, May 07 '12 at 16:58
The name of an array evaluates to the address of the array *except* when passed to the `sizeof` or address-of (unary `*`) operators. — Jerry Coffin, May 07 '12 at 17:00
@cnicutar: Indeed, I just noticed that. Deleting (rather than simply closing) duplicates doesn't strike me as a good idea. — Oliver Charlesworth, May 07 '12 at 17:00

score 0 · Accepted Answer · answered May 07 '12 at 17:22

0

I am no compiler expert but during the compilation phase, the compiler performs something called semantic analysis. During this phase, type checking is done. sizeof is also a compile time operator ( barring VLArrays probably ) and during type checking the compiler determines that in main, x is an array and in doit function it is a pointer.

Its like, the compiler is the owner of the house and hence it knows the type of its tenants.

Read about the compilation process on wiki

answered May 07 '12 at 17:22

Pavan Manjunath

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The `doit` function parameter _is a pointer_. When an array is passed as a function parameter it always decays to a pointer. – Blastfurnace May 07 '12 at 17:37
@Blastfurnace I guess I haven't contradicted to the point you are making? May be the wording confuses you? Yes the function parameter is indeed a pointer as you cant pack arrays and send them – Pavan Manjunath May 07 '12 at 17:39
You kind of dance around the answer, but never really hit it. The issue is that when an expression of array type appears in most contexts (such as in a function call), it is converted ("decays") to an expression of pointer type. IOW, you have the cause and effect reversed. The conversion doesn't happen because `doit` expects a pointer; rather, `doit` expects a pointer because the conversion happens. – John Bode May 07 '12 at 18:37

sizeof operation on char array

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