警告：赋值使指针来自 integer 没有强制转换 [-Wint-conversion] p = (i + 1) * 100;

Question

I am trying to fill array with the numbers without using scanf . I encountered warning: assignment makes pointer from integer without a cast [-Wint-conversion]

p = (i + 1) * 100;
  ^

and when i try to print the array the array output random values. How do i solve it?

#include<stdio.h>

int main() {
    int nums[8], i;
    int *p;
    p = nums;
    for (int i = 0; i < 8; ++i)
    {
        p = (i + 1) * 100;
        p++;
    }

    return 0;
}

Answer 1

You declared a pointer

int *p;
p = nums;

and then instead of assigning the element of the array pointed to by the pointer you are trying to assign the pointer itself with an integer value.

p = (i + 1) * 100;

It is evident that you mean

*p = (i + 1) * 100;

Pay attention to that it is a bad idea to use the magic number 8 in the for loop. It is better to use a named constant.

If you want to fill the array using a pointer in the for loop then the program can look the following way

#include<stdio.h>

int main( void ) 
{
    enum { N = 8 };

    int nums[N];

    int init_value = 1;
    for ( int *p = nums; p != nums + N; ++p )
    {
        *p = 100 * init_value++;
    }

    return 0;
}

Using this approach you can write a separate function that will initialize an array the following way

void init_array( int *a, size_t n, int value )
{
    for ( int *p = a; p != a + n; ++p )
    {
        *p = 100 * value++;
    }
}

Answer 2

In the assignment p = (i + 1) * 100 , p is an int * , not an int , so the assignment would convert the integer (i + 1) * 100 to a pointer. Instead, use *p = (i + 1) * 100 .