Pointer in simple variable vs in array

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C

// CODE  1
#include <stdio.h>
int main()
{
    float cgpa[] = {25, 39, 45,95,98,9,
    9,48,45,546,8,41,4,8,84,84,95};


int *pointer = &cgpa[1];

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


      return 0;
}

Versus the below code:

C

//CODE 2
#include <stdio.h>
    int main(){ 

int a =5;
int *b=&a;
*b=8;
int d =5;
int *e= &d;
printf("The sum of a and b is %d\n", *e+*b);
printf("the addres a =%u\n",&a);
printf("the addres a =%u\n", b);
printf("the value  a =%d\n",*b);
printf("the value  a =%d\n",a);
printf("the value  b =%d\n",b);

return 0;
}

What I have tried:

I want to get the value in the variable using the pointer. In first code sample I tried to do it with array, but failed. In the second code sample I was able to execute. Why?

Posted 2-Jun-22 1:05am

Aayush Jain 2022

Updated 2-Jun-22 7:56am

Sean Ewington

v3

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OriginalGriff · Accepted Answer · 2022-06-02T01:40:00

FIrst off, arrays in C start from a zero index, not one.
So if an array arr has three elements 101, 102, and 103 then you can only access it's content using then indexes 0, 1, and 2:

arr[0] == 101
arr[1] == 102
arr[2] == 103

Second, integers and floats are not necessarily the same size: an int is commonly 16 bits (or 2 bytes) or 32 bits (or 4 bytes) wide, while a float is generally 32 bits (4 bytes) wide. Trying to use an int pointer to access float data may or may not works at all, depending on your compiler and / or operating system!

Thirdly, floats and integers aren't stored the same way: int is stored as a "straight binary value" with the most significant bit determining if the value is positive or negative. A float is much more complicated: IEEE 754 - Wikipedia[^]

So again, using a int pointer to access a float array isn't going to give you the results you expect!

Try it like this:

C

#include <stdio.h>

int main()
    {
    float cgpa[] = {25, 39, 45, 95, 98,  9,
                     9, 48, 45,546,  8, 41,
                     4,  8, 84, 84, 95};

    float *pointer = &cgpa[0];
    printf("%0.0f\n",*pointer);

    return 0;
    }

And it'll start to work.

But ... the name of an array is a pointer to it's first element (that's defined in the language specification) so this code is both more readable and simpler:

C

#include <stdio.h>

int main()
    {
    float cgpa[] = {25, 39, 45, 95, 98,  9,
                     9, 48, 45,546,  8, 41,
                     4,  8, 84, 84, 95};

    float *pointer = cgpa;
    printf("%0.0f\n",*pointer);

    return 0;
    }

Richard MacCutchan · Accepted Answer · 2022-06-02T01:51:00

Solution 2

Get yourself a copy of The C Programming Language (2nd Edition): Amazon.co.uk: Ritchie, Dennis, Kernighan, Brian: 8601410794231: Books[^] where pointers (and all the other concepts of C) are clearly explained.

Posted 2-Jun-22 1:51am

Richard MacCutchan