Basic C programming

#include<stdio.h>
void main()
{
printf("%d",printf("Computer"));
}



has a output = 8

Why ? please explain someone.

 

why are you making two calls to printf for? also you formating is returning the length of the text computer. Computer is 8 characters long. when using printf you need to use a format that goes with your data type. Your using decimal %d instead of string %s and making an additional call to printf as well. Try this

Code:

#include<stdio.h>
void main()
{ 
printf("%s","Computer");
}

additionally you need to use some sort of debugger or make a scanf reference to see the output.

Code:

#include<stdio.h>
void main()
{ 
printf("%s","Computer");
char i = scanf("i");
}

in practice you don't call scanf like that(without formatting) but it will exit the program one you press enter allowing you to check the output. I'd say use system("PAUSE"); but that is windows only.

 

The code is valid, the output should actually be "Computer8". 8 is returned by printf, which returns the number of characters it prints.

 

Why dont we use '&' in the case of strings in function scanf ??

e.g. :-


scanf("%s",date);

here date is a character array means string.
There is no & before date
This the part of a code.


Plz help

 

Because date is already a pointer.

 

Even I had the same question what vivek is asking to you? But, Now I got it that date is already a pointer then it should be use scanf("%s",date) like that. I am learning the C language, So this information will really be useful for me.

 

The ampersand symbol '&' otherwise known as the address operator takes in the address of a certain variable. the scanf function requires addresses so therefore the '&' is necessary for data types such as int, short, long, double, float, etc....the fact the date is already a pointer (variables that hold ONLY addresses), there is no need to use the '&' to retrieve the address since this is what they are already holding.