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problem with sizeof

hi
i wrote this code which is written bellow when ever i check the value of u it appear to be the 3. while it is supposed to be 8.
Can any body tell me what is the problem with this code and how can i get the exact size of the arrgument of func.

void main(){
  func("fara.txt");
}
void func(unsigned char arr[]){
int u;
u=sizeof(arr);
}

thanks
Regards
Farhan Arshad

Parents

0 Dan Henry over 21 years ago

The problem is not with sizeof itself, but that you are applying it to an unbounded array (arr[]). The way you have the 'arr' parameter declared, it degrades to a pointer which, being unqualified, is a generic pointer whose size is 3.

Compare this example:

void main() {
    func("fara.txt");
}

void func(unsigned char arr[]) {
    int u;
    u = strlen(arr);
}

with this example:

void main() {
    func("fara.txt", sizeof "fara.txt");
}

void func(unsigned char arr[], size_t len) {
    int u;
    u = len;
}

and with this example:

char str[] = "fara.txt";

void main() {
    func(str, sizeof str);
}

void func(unsigned char arr[], size_t len) {
    int u;
    u = len;
}

Reply

0 Dan Henry over 21 years ago

void main() {
    func("fara.txt");
}

void func(unsigned char arr[]) {
    int u;
    u = strlen(arr);
}

with this example:

void main() {
    func("fara.txt", sizeof "fara.txt");
}

void func(unsigned char arr[], size_t len) {
    int u;
    u = len;
}

and with this example:

char str[] = "fara.txt";

void main() {
    func(str, sizeof str);
}

void func(unsigned char arr[], size_t len) {
    int u;
    u = len;
}

Children

0 Dan Henry over 21 years ago in reply to Dan Henry

I should note that the value assigned to 'u' in example #1's func() is different than the value assigned in examples #2 and #3. Why is left as a simple exercise for the reader.
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