return makes integer from pointer without a cast [-Wint-conversion] return candidate

Question

I'm learning C, but I find those vague compiler error messages increasingly frustrating. Here's my code:

char getkey (int minimo, int maximo, int alphalen, int index, char alpha[])  
{  
    int cociente, residuo, cont;
    int i = 0;
    char cand[maximo+1];
    char candidate[maximo+1];

    while (index != 0)
    {
        cociente = index / alphalen;
        residuo = index%alphalen;
        cand[i] = residuo;
        index = cociente;
        i+=1;
    }

    for (cont=i-1; cont>=0; cont--)  
    {   
        int pos = cand [cont];
        candidate[i] = alpha[pos];      
    }
    return candidate;
}

This generates one warning:

return makes integer from pointer without a cast
- return candidate;

Can someone explain these warning?

Your function expects to return char, but are you returning char? — SandBag_1996, May 04 '16 at 20:15
Your function is declared as returning `char` but you are trying to return a `char *`. — Paul R, May 04 '16 at 20:15
candidate is a char array, not a char and your function is declared so that the return value is expected to be that of a char. — Mike -- No longer here, May 04 '16 at 20:18

Steve Summit · Accepted Answer · 2016-05-04T21:00:58.513

7

Your local variable candidate is an array of char. When you say

return candidate;

you return a pointer to char. (This is due to the very close relationship between arrays and pointers in C, which you will have to learn about if you don't know yet.) But you've stated that your function getkey returns a char, not a pointer-to-char.

You need to make the function's return value match its type. You might want change the return statement so it returns a single char. If you want to have the function return a whole string, you could change its declaration to char *getkey(), but in that case you'll also have to take care of the allocation of the candidate array.

Here's what the warning means. You tried to return a pointer. The function is supposed to return a character, which is represented as a small integer. The compile is willing to try to convert the pointer to an integer, in case that's what you really want to do, but since that's not usually a good idea, it warns you. Indeed, in this case the warning tells you that you probably made a mistake.

edited May 04 '16 at 21:00

answered May 04 '16 at 20:16

Steve Summit

45,437
7
70
103

4

Is it really known as "equivalence of arrays and pointers in C"? Because that's not true. Arrays can decay into pointers under certain contexts (i.e. passing and returning) but they are not pointers. – Kevin May 04 '16 at 20:17
@Kevin "Equivalence" is what we used to call it. (Does nobody use that term any more?) "Equivalence" does not mean they're the same thing; just that they're closely -- even intimately -- related. See http://c-faq.com/aryptr/aryptrequiv.html . – Steve Summit May 04 '16 at 20:23
No, arrays and pointers are not closely related. One is a memory address, the other is _n_ elements laid out contiguously in memory. – Emil Laine May 04 '16 at 20:26
@tuple_cat "If arrays and pointers are not closely related", asks user5672720, "then when I tried to return an array from my function, why did it return a pointer instead?" – Steve Summit May 04 '16 at 20:32
@Adam Thanks, didn't think about that in this case. Edited some more. – Steve Summit May 04 '16 at 20:49
@SteveSummit: Actually "we" never used the term "equivalent", because it is plain wrong: `sizeof(int [10])` and `sizeof(&(int[10]))`) are not equivalent. But your rephreased text is fine. – too honest for this site May 04 '16 at 20:58
Thanks Steve, that looks much better. – Adam May 04 '16 at 21:01

score 2 · Answer 2 · edited May 23 '17 at 12:23

getkey has been declared to return a char, but you're trying to return candidate which is a char array.

You probably meant to return one of the elements in the array, i.e. something like candidate[x].

The reason for the error message saying "pointer" instead of "array" is that the compiler is converting candidate to a pointer to the first element in the array, which happens implicitly in some contexts, for example when returning, since arrays can't be returned (or passed as parameters) by value in C (see What is array decaying?).

return makes integer from pointer without a cast [-Wint-conversion] return candidate

2 Answers2

Linked

Related