8 replies to this topic

[Nyfer]

hey guys please help me in this

in the below code

#include<stdio.h>


void main()

{

float a=10.5;

float x=20.78;

printf("a=%d",a);

printf("x=%d",x);

getch();

}

am getting a=0 and garbage value for x. why is that can anyone please explain me
thanks

[Ancient Dragon]

use %f in the printf() format specifier, not %d, which is for integers.

[Nyfer]

ya i know that, but can you explain me why am getting a=0 and x=37833827 value??

[voral]

Because the variables of different types have different representation in memory, and can also vary in size.
This function does not convert the input parameters, and reads as an int.
Your code is not equivalent to "printf("x=%d",(int)x);"

[Flying Dutchman]

floating point data types store information differently than integer ones. Since, as far as I know, C can't tell of which type a variable is (in va_list, which is used in printf function), the function figures it out by looking at the format specifier.

[lespauled]

It may become clearer if you run this:

#include <iostream>

#include <stdio.h>


using namespace std;


int main()

{

    float a=10.5;

    float x=20.78;


    printf("Size of int....: %d\n", sizeof(int));

    printf("Size of double.: %d\n\n", sizeof(double));


    printf("a=%d\n",a);

    printf("a=%f\n",a);

    printf("x=%d\n",x);

    printf("x=%f\n",x);

    std:getchar();


    return 0;

}

[Nyfer]

sizeof int and float is same.
can you please explain me

[lespauled]

My bad, I put double. The problem is not the size of the variable in memory, the problem is how printf handles variables. Since printf does not do any conversions on it's own, the results are based on what you send to it in the text. The results might possibly be different on different machines. In memory, the representation for a float looks nothing like the representation for a signed int.

[Nyfer]

ok thank you:)