C Programming - Control Instructions - Discussion

Do it manually for a shorter number, ex while(j <= 3), you will get the answer as 3. Similarly when while(j <= 255), Answer is 255.

Since the int is 2 bytes here, wouldn't it become zero and the loop continue indefinitely. Please Help to clear this doubt.

%c will print the character according to ASCII values and %d print the corresponding number.

Do it for j<=60, you will find difference. Actually %c shows ascii characters and %d is a decimal value of it. But I am still confused for j<=255, whether this loop will be executed infinite or 255 times only.

I can't understand because of %c and j two times print.

#include<stdio.h>
int main()
{
char result[50];
float num = 23.34;
sprintf(result, "%f", num);
printf("\n The string for the num is %s", result);
getchar();
}

This program shows an error output too large. Can anyone tell me why this error is occuring?

@ Rahul & Nantu.

Your program will execute without error if you change like below,

#include<stdio.h>
#include<string.h>
void main()
{
char result[50];
float num = 23.34;
sprintf(result, "%f", num);
printf("\n The string for the num is %s", result);
getchar();
}

In your program you wanted to convert the float value to string [sprintf(result, "%f", num);]

So you have to include "string.h" header file.

If you are not give return type [return 0;] means you should mention void in main function instead of int main.

Hope you understand.

Hai friends some of you are telling the answer as 225 how you people got that I can't will you explain me. Because am confused in that.

Anyone help me for that.

What is the use of and meaning of "short int" here can anybody explain me?

Short int is the signed int and it usually used as "int".

Default of system declare "int" has 4 byte. But the question declare short int has 2 byte wide, so the range from: -2^8 -> 2^8 - 1.

This mean j only from: -256 -> 255.

So the while loop will be repeat infinite times. Be cause it's never has value 256 to break the loop.

When counter from j = 1 to j=255; j++ will be: j = -256, and the loop continue. Program will never be exit.

I'm got this experience when I'm test Pascal program in high school. So please think about this carefully.