#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
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.