C Programming - Input / Output - Discussion

Still I am facing problem in input and output statements.

OK ! I can explain it in a easier way:

Consider the statement:-

printf("abc1 %c %s %c abc2", 10, xyz, 10);

You can see its a simple printf statement.

Now replace, abc1 => char *str =

abc2 =>; main(){printf(str, 34, str, 34);}

10=> 34.

xyz =>char*str = %c%s%c; main(){printf (str, 34, str, 34);}

You will get the answer.

What is mean by char*string;? Could you explain please?

How could understand in easiest way?

"=asci value of 34.

Then str is? How and what will be correct?

Why once again main() printf(str,34,str,34) is printed at last? I don't understand. Please can anyone explain this code?

@Nivethitha

char *str = "char *str = %c%s%c; main(){ printf(str, 34, str, 34);}";

Given print stmt: printf(str, 34, str, 34);

Here,

The first str in printf is replaced by "char *str = %c%s%c; main(){ printf(str, 34, str, 34);}";

And the remaining arguments are as such, thus it will now become

printf("char *str = %c%s%c; main(){ printf(str, 34, str, 34);}" , 34,str,34);

The first character is " so the contents will be starting printing to output..while printing
the escape sequence %c is replaced by ASCII value of 34 which is "

Then %s is replaced by str which is "char *str = %c%s%c; main(){ printf(str, 34, str, 34);}"--the escape sequence here are not substituted bcoz we have not specified any variables for it..

%c is again replaced by ASCII value of 34 which is "

Thus output is:
char *str = "char *str = %c%s%c; main(){ printf(str, 34, str, 34);}"; main(){ printf(str, 34, str, 34);}

Thank you @Vishnupriya.

Thanks @Vishnupriya.

I'm not getting this.

Please give a proper explanation.

Awesome explanation, thank you all.