C Programming - Control Instructions - Discussion

printf always return no of character.

Do that you can understand better.

printf("%d",printf("%d",printf("Hi how are you"))).

Here it is like;
If(pf=( ) ) means 0,
So, ch=0
If(0) it means condition is false.
So, else is printed.

When I use:

#include<stdio.h>
int main()
{
char ch;
if(ch == printf(""))
printf("It matters\n");
else
printf("It doesn't matters\n");
return 0;
}

Why does it print it matters?

if( ch = 0) // it's true
In case ch = 5
if (ch = 0) // this also true Statement because here we used single equal.

In printf (" "); there is a space so it takes space as a character and assigns value ch=1.

So, the output will be It matters.

#include<stdio.h>
int main()
{
char ch;
if(ch = printf("Hai"))
printf("\nIt matters\n");
else
printf("It doesn't matters\n");
return 0;
}

It is executed and prints "Hai" follows "It matters", character datatype assigns only one character as well as if condition checking only for '==' how will works?.

Hey I have a doubt here in printf (==) if it is there it works. So the above question doesn't works.

Here, printf() is a function that is used to evaluate the condition in if statement. We need to check whether printf() is printing anything.

If yes, then the If statement will evaluate to true. If it does not print anything then the if statement evaluate to false.

In this program printf() function doesn't print anything, so the condition is false and hence the result "It doesn't matters". Also the variable being char or integer doesn't matter.

if(ch=printf(""))
printf("matters");
else
printf("it does not matters");

The above snippet will not execute in Turbo C, it will show error IF statement, its incorrect statement.

Error: "possibly in correct statement".

printf("") returns 0 to any variable then
ch=0
if (ch)means if(ch!=0) which is not true anytime so else expression will be printed