Basic Concepts of C-Programming

Question

#include<stdio.h>
main()
{
    char *s="Manikant";
    int i=0;
    printf("%s",s+s[2]-s[4]);  // Line-1
    printf("\n%d",printf("GATE-2017"));  //Line-2
    printf("\n%c%d",s[i++],i);    //Line-3
}

What is output of Line-1 , Line-2 and Line-3 ?

Answer 1

#include<stdio.h>
int main() {
    char *s="Manikant";
    int i=0;
    printf("%s\n",s+s[2]-s[4]);  // Line-1
    printf("%d\n",printf("GATE-2017"));  //Line-2
    printf("%c%d\n",s[i++],i);    //Line-3
}

O/P

ikant        
GATE-20179
M1 // undefined [sequence point error]

s+s[2]-s[4] evalutes to s+3

printf("GATE-2017") , here 9 characters are printed by inside printf(). So inside printf() returns 9. => first printf prints 9 after GATE-2017

S[i++] expression evalutes to S[0] because of post increment.

Line3:

If an object is written to within a full expression, any and all accesses to it within the same expression must be directly involved in the computation of the value to be written.

printf("%c%d\n",s[i++],i); Here, we are accesing $i$ in the second argument of printf() but, it has nothing to do with the storing of $i$,which is a side effect of s[i++]. => causes undefined behaviour. Link

Comments

i stil didn't get :(

str , str[2], are holding address , right ?
*str , *str[2] contains letters.

substracting str[2]-str[4]  // this is subtracting addresses .,right ?

---

str is a character pointer and is holding first address of string literal "helloworld".

Now, str[x] = *(str+x) = *(x+str) (assume x > 0)

=> start from the memory location pointed by str and go x steps forward, once you get there, dereference it, collect the character in that location.

=> str[x] : is a character.

---

And what Sourav told is important. Because ASCII code is not part of standard C. (C might use someother character code as per the ssytem). But C standard mandate that alphabets must be coded with consecutive integers in same sequence.

Answer 2

$s\left [ 2 \right ]=n,\, s[4]=k$

Let ASCII value of $s\left [ 4 \right ]=k=X$$\Rightarrow s[2]=k+3$

 printf("%s",s+s[2]-s[4]); 

  printf("%s",s+3);

it will print from s[3]to end where it finds NULL,hence output$\Rightarrow ikant$

 printf("%d\n",printf("GATE-2017"));$\Rightarrow$printf returns number of characters it printed!

Internal Printf will run 1 time and will print"GATE2017" and then 9 (number of characters it printed returned by printf)

hence output -:GATE20179

 printf("%c%d\n",s[i++],i);

It will print s[0]=m and then i=1 hence output will be 

m1