What is the difference between char[1000] and int[1000]

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Intelligent Girl | Introduction to Dynamic Programming 1 & Algorithms Practice Problems | HackerEarth[^]

its a ques from hackerearth. i have submitted successfully. but i have some doubts.

What I have tried:

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

    int i,count,j,l;
    char arr[10000];
    scanf("%s",arr);
    
    for(i=0;i<strlen(arr);i++)
    {
            count=0;
    for(j=i;j<strlen(arr);j++)
    {

        if(arr[j]%2==0)
        {

            count++;
        }
    }
        printf("%d ",count);


    }
    return 0;
}

if i will write int arr[10000] instead of char[10000]. it is not working plzz explain me this.

Posted 1-Nov-17 21:34pm

jatinp510

Updated 2-Nov-17 1:27am

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4 solutions

Solution 4

Not your question, but your program can be made faster.
- Odd/even
Using Modulo to know if a number is odd or even, is more or less doing a division, it cost a lot.
Since the least significant bit is the answer, you can get the answer with a simple bitwise and
- get rid off the nested loops
When you compare 2 consecutive results, they differ only by the odd/even result of first element in sequence.
So once you know the result of full sequence, the second result depend only on the odd/even status of first element, you don't need to recalc the whole sequence.

Posted 2-Nov-17 1:27am

Patrice T

Updated 2-Nov-17 12:15pm

v2

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OriginalGriff · Accepted Answer · 2017-11-01T22:02:00

char and int are different sizes, and the system is well aware of this. char variables occupy a single byte, while int values occupy 4 these days - assuming a 32 bit integer, which is normal these days.
When you declare arr as an array of char and use scanf to read a string into it, it reads it fine, and your array indexes fetch a char from each byte - as the index increases by one each time, so does the memory address from which each is fetched.

----- arr points here
|
v
abcdefghijklm
^^^
|||
||--- arr[2]
||
|---- arr[1]
|
----- arr[0]

When you declare it as an int array, the scanf does exactly the same thing, but the indexes increase the memory address by the size of an integer - 4 bytes, rather than one:

----------- arr points here
|
v
abcdefghijklm
^   ^   ^
|   |   |
|   |   --- arr[2]
|   |
|   ------- arr[1]
|
----------- arr[0]

Afzaal Ahmad Zeeshan · Accepted Answer · 2017-11-01T21:53:00

There is a data type mismatch, although mostly it does not prevent you from running the program, only it will give you wrong outputs.

But you need to understand the basic difference in int type and char type, especially in C context. In high-level languages, they sometimes have a same size for data. In C, they have different (a lot different) sizes, char is only 1 byte, and int is 4 bytes (on 32 bit, and different on others). C data types - Wikipedia[^]

Later on, you are trying to do a %d, which expects a decimal. And same is the case for the %s, which expects a char type. So you are printing in one format, and accepting in another, which C will never complain about — only you will. You need to change either %d and work in all characters, or change that %s to work with integers only. Also, once you have made this change, you can select what type of array you want to work with.

Consider reading here, scanf format string - Wikipedia[^]

CPallini · Accepted Answer · 2017-11-01T22:02:00

Solution 2

Difference: 30000 bytes of memory.

In C programming language, A char is a 1-byte integer while a int is larger (typically 4 bytes).
Strings are array of chars (i.e. functions handling strings, like scanf with %s format specifier, and strlen do expect array of characters as arguments).
You are going to get compiler warnings (and incorrect behaviour) if you use an array of int in your program.

Posted 1-Nov-17 22:02pm

CPallini

Updated 1-Nov-17 22:09pm

v2

Comments

jatinp510 2-Nov-17 4:20am

thank u..

CPallini 2-Nov-17 4:24am

You are welcome.

What is the difference between char[1000] and int[1000]

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