Print elements of array of void pointers

[Shervan360]

Hello,

How can I fix this? I want to print all data in this array.

# include <stdio.h>



int main()
{
	void* myPointer[4];
	int a = 10; float b = 30; char c = 'C'; double d = 1.5;

	myPointer[0] = &a;
	myPointer[1] = &b;
	myPointer[2] = &c;
	myPointer[3] = &d;

	printf_s("%d\n", *(int*)myPointer);
	printf_s("%f\n", *(float*)(myPointer+1));
	printf_s("%c\n", *(char*)(myPointer + 2));
	printf_s("%ld\n", *(double*)myPointer + 3);


	return 0;
}

[Ganado]

What is the type of (int*)*myPointer?

What does the printf format specifier "%d" expect it to be?

[Shervan360]

Thank you @Ganado,
I edited pointers in printf.

I want to cast like this.

#include<stdlib.h>
int main() {
   int a = 7;
   float b = 7.6;
   void *p;
   p = &a;
   printf("Integer variable is = %d", *( (int*) p) );
   p = &b;
   printf("\nFloat variable is = %f", *( (float*) p) );
   return 0;
}

[Aioria]

hello, be guided by this example

#include <iostream>

using namespace std;

int main()
{
    int pointer[3][2] = {{1,2},{9,8},{14,21}};
    int fil = (sizeof(pointer)/sizeof(pointer[0]));
    int col = (sizeof(pointer[0])/sizeof(pointer[0][0]));
    for (int i = 0; i < fil; i++)
    {
        for (int j = 0; j < col; j++)
        {
            cout<<pointer[i][j]<<endl;
        }
    }
}

[Handy Andy]

Hello Shervan360,

# include <stdio.h>

int main()
{
	void* myPointer[4];
	int a = 10; float b = 30; char c = 'C'; double d = 1.5;

	myPointer[0] = &a;
	myPointer[1] = &b;
	myPointer[2] = &c;
	myPointer[3] = &d;

	printf("%d\n", *(int*)*myPointer);  // Added "8" before "myPointer".
	printf("%f\n", *(float*)*(myPointer+1));
	printf("%c\n", *(char*)*(myPointer + 2));
	printf("%g\n", *(double*)*(myPointer + 3));  // <--- Should match the 2 previous lines.

	return 0;
}

Andy