错误：语义问题从不兼容类型“ void”分配给“ int”

Question

While creating a program with function prototype, a problem has occurred. It said:

Semantic issue Assigning to 'int' from incompatible type 'void'.

Could you please help me resolve this issue?

Here is my code:

#include <stdio.h>
#include <math.h>

void powr(int);

int main(void) {

    int n=1, sq, cu, quart, quint;

    printf("Integer  Square  Cube  Quartic  Quintic\n");

    do {

        sq = powr(n); //here is the error line
        cu = powr(n); //here is the error line
        quart = powr(n); //here is the error line
        quint = powr(n); //here is the error line
        printf("%d  %d  %d  %d  %d\n", n, sq, cu, quart, quint);
        n++;
    }
    while (n<=25);

    return 0;
}

void powr(int n)
{
    int a, cu, quart, quint;

    a=pow(n,2);
    cu=pow(n,3);
    quart=pow(n,4);
    quint=pow(n,2);
}

Answer 1

void powr(int n)

means that the function will return nothing, so you're not allowed to do something like:

sq = powr(n);

If you want your function to take an int and return an int , it should be:

int powr(int n)

(for both the prototype and the function definition).

---

In any case, variables that you set within the powr function are not available to the caller (and using globals is a very bad idea in general), so you'd need to either change the function to return just the square of the number and call it thus:

sq = powr (n);
cu = n * sq;
quart = powr (sq);
quint = n * quart;

Or you could pass the addresses of the variables into the function so they could be changed, something like:

void powr(int n, int *pSq, int *pCu, int *pTo4, int *pTo5) {
    *pSq = pow (n, 2);
    *pCu = *pSq * n;
    *pTo4 = pow (*pSq, 2);
    *pTo5 = *pCu * *pSq;
}

and call it with:

powr (n, &sq, &cu, &quart, &quint);

I would suggest using the former approach, given the level you appear to be learning at (no offence intended, just stating that to assist you in selecting the appropriate method).