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Home/ Questions/Q 7756447
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Editorial Team
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Editorial Team
Asked: June 1, 20262026-06-01T12:49:36+00:00 2026-06-01T12:49:36+00:00

I want to calculate the ‘sizeof’ array: char* arr[] = { abc, def };

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I want to calculate the ‘sizeof’ array:

char* arr[] = { "abc", "def" };

When I call sizeof manually, immediately after the initialization of the array, it works fine. However if I pass the array to some function, It doesn’t give the same result.

int test(char* b[]) {
    return (int)sizeof(b);
}

int _tmain(int argc, _TCHAR* argv[])
{
    char* arr[] = { "abc", "def" };
    int p = test(arr); // gives 4
    int k = sizeof(arr); // gives 8
    ...
}

So what’s the problem? Sorry for the newbie question, but I really miss it.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-01T12:49:37+00:00Added an answer on June 1, 2026 at 12:49 pm

    When you use sizeof on arr in _tmain, you’re taking the size of the array itself, which is 2 * sizeof(char*) = 2 * 4 = 8 on your architecture, because your char*s occupy 32 bits = 4 bytes.

    When you use sizeof on b in test, you’re taking the size of a pointer to the first element of the array, which is sizeof(char**) = 4 on your architecture. The reason you’re taking the size of the pointer and not the array is that you can’t pass arrays as such to functions – instead, you pass them via a pointer to their first element. The information about the actual size of the array is thus lost in the call. Put another way, the signature of your function test is equivalent to int test(char **b).

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