char pointer as function parameter

问题

I have a function A that has a char * parameter. In A(), I use sizeof to calculate the size of buf , but I got 8 in i686 machine and 4 in armel machine. why is not it 128? The snippet is as below:

void A(char *p)
{
    printf("sizeof p is %d\n", sizeof(p));
}

int main(void)
{
    char buf[128];
    printf("sizeof buf is %d\n", sizeof(buf));
    A(buf);
    return 0;
}

the result is like this(in i686):

sizeof buf is 128
sizeof p is 8

Please tell me the reason.

回答1:

Because you print the size of the pointer, and not what it points to. Arrays decays to pointers, and as soon as you pass it to a function you have a pointer instead, and loose all size information of the original array.

回答2:

In main() function, buf is an array, so sizeof buf gives the size of this array: sizeof(char[128]) (128 bytes). That's because an array is not converted to a pointer when used as operand of sizeof operator.
In A() function, p is a pointer, so sizeof p gives the size of this pointer: sizeof(char *) (8 bytes on your implementation).

回答3:

sizeof(buf) is 128*sizeof(buf[0]) and sizeof(p) is just sizeof(char*).

回答4:

When passed into function, it decays to an ordinary pointer that has size of your native size(4 or 8).

Use strlen() because it checks for a "null" at the end and gives correct size if the array is already initalized.

来源：https://stackoverflow.com/questions/17424078/char-pointer-as-function-parameter

标签

sizeof