arrays - Why does the program display warning passing argument 1 of 'circularswap' makes pointer from integer without a cast?

Question

I am working on this problem:

Given an array p[5], write a function to shift it circularly left by two positions. Thus, if p[0] = 15, p[1]= 30, p[2] = 28, p[3]= 19 and p[4] = 61 then after the shift p[0] = 28, p[1] = 19, p[2] = 61, p[3] = 15 and p[4] = 30. Call this function for a (4 x 5 ) matrix and get its rows left shifted.

Here's the code I tried:

    #include <stdio.h>
    
#include <stdio.h>
void circularswap(int arr[][5],int n,int m){ int arr1[n][m],i,j;
    for (i=0;i<n;i++){
        for (j=0;j<m;j++){
            arr1[i][j]=*(*(arr+i)+j);
        }
    }
    for (i=0;i<m;i++){
        *(*(arr+i)+0)=arr1[i][2];
        *(*(arr+i)+1)=arr1[i][3];
        *(*(arr+i)+2)=arr1[i][4];
        *(*(arr+i)+3)=arr1[i][0];
        *(*(arr+i)+4)=arr1[i][1];
    }
        for (i=0;i<4;i++){
        for (j=0;j<5;j++){
            printf ("%d",arr[i][j]);
        }
        printf ("
");
    }
}
int main(){ int i,j;
    int arr[4][5]={(15,30,28,19,61),(15,30,28,19,61),(15,30,28,19,61),(15,30,28,19,61)};
    circularswap((arr,4,5));
    return 0;
}

Unfortunately, this shows up warnings. Can someone please tell why the warnings pop up and how to remove them?

question from:https://stackoverflow.com/questions/65880514/why-does-the-program-display-warning-passing-argument-1-of-circularswap-makes

Answer 1

You have to many parentheses when calling the function.

The expression (arr,4,5) is using the comma operator and will evaluate all the sub-expressions in the list, but return only the 5.

I.e. your call is really the same as circularswap(5), which is incorrect in multiple ways.

To solve your problem drop the inner parentheses:

circularswap(arr,4,5);

---

You have a similar problem when initializing your array: You use parentheses () instead of curly braces {}.

So the array definition is really equal to:

int arr[4][5]={ { 61 }, { 61 }, { 61 }, { 61 } };

_{[Note how I use curly-braces in the example above]}

---

On another note, for any pointer p and index i, the expression *(p + i) is exactly equal to p[i]. The latter is easier to read and understand, and also less to write.

It matters especially when using arrays of arrays like you do, where e.g. *(*(arr+i)+0) could be replaced with arr[i][0].