CODEWITHSUNDEEP
carrayssortingsizeof
cjstineb
cjstineb

Reputation: 17

Why does sizeof() not return the length of an array?

#include <stdio.h>

int main() {
    int test[3];
    int i;
    test[0]=5;
    test[1]=10;
    test[2]=7;

    printf("array size: %d\n",sizeof(test));
    sortArray(test);

    for(i=0;i<sizeof(test);i++) {
        printf(" %d ", test[i]);
    }
    printf("\n");
}

void sortArray(int number[]) {
    int i,j,a;
    int n = 5;

    for (i = 0; i < n; i++) {
        for (j = i + 1; j < n; j++) {
            if (number[j] < number[i]) {
                a = number[i];
                number[i] = number[j];
                number[j] = a;
            }
        }
    }
}

The array I am having problems with is "test" When I run the program the "size" is always a multiple of 4 of the intended size. ex: test[3] would output a size of 12. What am I doing wrong? I am using code::blocks as an ide also.

Upvotes: 1

Views: 304

Answers (1)

Foggzie
Foggzie

Reputation: 9821

sizeof returns the memory size of what you pass it. The return value is...

measured in the number of char-sized storage units required for the type

In a typical 32-bit system, a char is one byte and int is four bytes so you'll get a multiple of four for an array of type int.

If you'd like the length of the array, just divide by the size of the type:

int a[3];
size_t n = sizeof(a) / sizeof(a[0]);

Note: As dbush mentioned in the comments below:

...this only works if the array is not a paramerer to a function. In that case the array decays to a pointer and sizeof(array) evaluates to the size of a pointer.

Upvotes: 5

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