How can I find the sum with the pointers?-CodePudding

How can I fix that with pointers ? with *p ? How can ı find the total of columns? I want to choose which column ı chose and see the total of it. lets say array is this:

then here is my program fragment:

for(*p=0;p<size;p  )
sum=sum *p;
printf("\n");

CodePudding user response：

Using pointers:

#include <stdio.h>

#define Size(x) (sizeof(x) / sizeof(*x))

int main() {
    int array[][3] = {
        {2, 6, 9},
        {5, 6, 9}, 
        {4, 8, 4},
        {2, 6, 0},
        {4, 6, 7}
    };

    int selected_column;
    // Size(*array) will be 3, the number of columns, that is:
    //     sizeof(int[3]) / sizeof(int)
    printf("Select column (1 - %zu): ", Size(*array));
    if(scanf(" %d", &selected_column) != 1 ||
       selected_column < 1 || selected_column > Size(*array)) return 1;
    --selected_column; // make it zero-based

    int sum = 0;
    // Size(array) will be 5, the number of rows, that is:
    //     sizeof(int[5][3]) / sizeof(int[3])
    for(int(*p)[Size(*array)] = array, (*e)[Size(*array)] = p   Size(array);
        p != e;   p) 
    {
        sum  = (*p)[selected_column];
    }

    printf("col %d sum = %d\n", selected_column, sum);
}

A much simpler and idiomatic approach would be to iterate over he outer array (consisting of int[3]s):

#include <stdio.h>

#define Size(x) (sizeof(x) / sizeof(*x))

int main() {
    int array[][3] = {
        {2, 6, 9},
        {5, 6, 9}, 
        {4, 8, 4},
        {2, 6, 0},
        {4, 6, 7}
    };
    
    int selected_column;
    printf("Select column (1 - %zu): ", Size(*array));
    if(scanf(" %d", &selected_column) != 1 ||
       selected_column < 1 || selected_column > Size(*array)) return 1;
    --selected_column;

    int sum = 0;

    for(int row = 0; row < Size(array);   row) {
        sum  = array[row][selected_column];
    }

    printf("col %d sum = %d\n", selected_column, sum);
}

Output from both versions:

col 0 sum = 17

CodePudding user response：

Your code should look like this if you want to do it with a pointer.

#include<stdio.h> 

int main(){ 

int arr[5][3] = {{2, 6, 9},{5, 6, 9},{4, 8, 4},{2, 6, 0},{4, 6, 7}};

int sum = 0;
int *p = arr;

for(int i=0;i<3;i  ){
    sum = sum   *(p i);
}

printf("%d\n",sum);

return 0; 
}