array pointer basic

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

Code:

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#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;
}

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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

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().

Quote:

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Originally Posted by 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().

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line 12

Code:

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pointer = &array[0];

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problem with this line

Code:

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

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I tried with different approach