在 C99 的除法中使用指针，错误：二进制操作数无效

Question

The question surely look stupid, but I have always wasted a lot of time, with tests/errors until it works, with this kind of problem.

I need a pointer to return a value from a function, then I need to divide this value, but I have the compiler error:

invalid operands to binary / (have 'int *' and 'int')

Example in the following code:

void test(int *x, int y)
{
    *x = *x+y;
}

int main()
{
    int *x = 20;
    int y = 1;

    test(&x, y);

    printf("x: %i", x);

    float res = x / 2; // error: invalid operands to binary / (have 'int *' and 'int')

    //float res = *x / 2; // application crash with this one. (GCC native Android)
}

I tried with '&' but it didn't work (x = 0), I try cast to int and only get the pointer adress, etc.

Answer 1

Your immediate problem is that you have declared x as a pointer to int, but you are trying to use it as an int. To simply correct the last line, dereference the pointer (just like you've correctly done inside test() ):

float res = *x / 2;

Now, it appears you actually tried that and got an error, which is not surprising, because you've initialized x badly:

int *x = 20;

This doesn't create an int value of 20 and make x point to it. It sets x to point at the memory address represented by the integer value 20. That's probably reserved memory, which is why you get an error when you try to dereference it.

(You don't get that error in test() because you've passed the address of x as the argument. So dereferencing it there actually does get you 20 - probably.)

To make the pointers work, either do:

int x = 20;
...
test(&x, y)
...
float res = x / 2;

or:

int *x = malloc(sizeof(int));
*x = 20;
...
test(x, y)
..
float res = *x /2;

But you are really making this too difficult. Since you only need to output one value from the function, just make the function return that value. Then you have no need to mess about with pointers at all:

int test(x,y) { return x+y; }
...
int x = 20;
...
x = test(x,y);
...
float res = x / 2;

(And finally, I believe that in any case you want to use 2.0 in the last line, not just 2 , if you want to get a float result instead of an int.)

Answer 2

int * x = 20; means: x is a pointer to an int stored at address 20 . When you deference it, the CPU will try to read an int value from address 20 , yet address 20 is an invalid address in your current process space and thus your program crashes.

What you probably wanted do say is: x is a pointer to an int and the value of the int it points to is 20 , correct? Well, this would have been:

int someInt = 20;
int * x = &someInt;

Now x points to whatever address someInt stores its value (actually it's on the stack space of the main function, but that's just a side node) and the value stored at that address is in fact 20 .