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Home/ Questions/Q 861749
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T09:01:45+00:00 2026-05-15T09:01:45+00:00

If I can pass in an array of a known size: void fn(int(*intArray)[4]) {

  • 0

If I can pass in an array of a known size:

void fn(int(*intArray)[4])
{
    (*intArray)[0] = 7;
}

why can’t I return one:

int intArray[4] = {0};
int(*)[4] fn()
{
    return &intArray;
}

here, the “)” in “(*)” generates “syntax error : )”.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-15T09:01:46+00:00Added an answer on May 15, 2026 at 9:01 am

    The [4] goes after the function name, just like it goes after the variable name in a variable definition:

    int (*fn())[4]
    {
        return &intArray;
    }
    

    Since this is very obscure syntax, prone to be confusing to everybody who reads it, I would recommend to return the array as a simple int*, if you don’t have any special reason why it has to be a pointer-to-array.

    You could also simplify the function definition with a typedef:

    typedef int intarray_t[4];
    
    intarray_t* fn() { ... }
    
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