C string array sizeof() changes

Question

void func(char *s[]){
    printf("s: %d\n", sizeof(s));
}

void caller(){
    char *a[2];
    for(int i = 0; i < 2; i++){
        a[i] = (char *)malloc(50 * sizeof(char));
    }
    strcpy(a[0], "something");
    strcpy(a[1], "somethingelse");

    printf("a: %d\n", sizeof(a));

    func(a);
}

This outputs

a: 16
s: 8

Why is the output of sizeof() different in caller() and func()? Also, is it possible for func to get the number of char * in the array via sizeof?

Welcome to strings/arrays in C. It's decaying to a pointer. – Richard J. Ross III Mar 26 '13 at 17:58 — Richard J. Ross III, Mar 26 '13 at 17:58

score 0 · Answer 1 · answered Mar 26 '13 at 18:02

0

One is an array of two character pointers. The other is a pointer to an array of pointers. They're not the same size.

answered Mar 26 '13 at 18:02

John

15,744
9
65
107

Was it really necessary to answer this instead of voting to close? This has been answered many times before, and should be closed. – Richard J. Ross III Mar 26 '13 at 18:02
I wanted to try to answer it. I'm not interested in checking to see if a question has been answered before I try to answer it. I'm here to learn by testing my understanding. If that's annoying to you, I'm sorry. – John Mar 26 '13 at 18:05

Ritesh Kumar Gupta · Answer 2 · 2013-03-26T18:10:27.390

0

In C/C++,Array decays into a pointer.char *s[] is a pointer, the size of pointer is always fixed 4 or 8 (depends upon machine).

edited Mar 26 '13 at 18:10

answered Mar 26 '13 at 18:03

Ritesh Kumar Gupta

4,525
6
39
68

Actually, no. It's decaying to a pointer in this scenario. – Richard J. Ross III Mar 26 '13 at 18:04
@RichardJ.RossIII :Exactly decaying pointers.I forgot the exact name.I Updated my answer. Thanks! – Ritesh Kumar Gupta Mar 26 '13 at 18:06

C string array sizeof() changes

2 Answers2