array pointer basic

[abhi143]

I believe value of pointer should be equal to address of array element

#include<stdio.h>

int main (void)


{
    int i;
    
    int *pointer;
    
    int array[5] = {10, 20, 30, 40, 50};
    pointer = &array;
    
    for (i = 0; i <5; i++)
    {
           printf("Memory Allocation %p : [%d]   = %d \n", &array[i], i, array[i] );
    }
    
    for (i = 0; i <5; i++)
    {
        printf("%p : %d \n", (pointer++), (*pointer++));
    }
    
    
    printf("size of array : %d \n", sizeof(array));
    
    return 0;
}

Memory Allocation 0061FF14 : [0] = 10
Memory Allocation 0061FF18 : [1] = 20
Memory Allocation 0061FF1C : [2] = 30
Memory Allocation 0061FF20 : [3] = 40
Memory Allocation 0061FF24 : [4] = 50

0061FF18 : 10
0061FF20 : 30
0061FF28 : 50
0061FF30 : 3
0061FF38 : 2404352

size of array : 20

address of array should be same as the pointer value

In my program I am not getting output as I want, result show I am doing mistake. What's wrong with code

[jimblumberg]

A couple of things.

First line 12 is incorrect (look at line 16 for a hint).

Second line 21 is invoking undefined behaviour, you're attempting to increment the same value twice without a sequence point. You should probably move the increment to a separate line after the printf().

[abhi143]

A couple of things.

First line 12 is incorrect (look at line 16 for a hint).

Second line 21 is invoking undefined behaviour, you're attempting to increment the same value twice without a sequence point. You should probably move the increment to a separate line after the printf().

line 12

 pointer = &array[0];

problem with this line

printf("%p : %d \n", (pointer++), (*pointer++));

I tried with different approach