C Programming

Question

Explain the Output of the Following C Code:

 

int main()
{
char s[]={'g','a','t','e','c','s'};
char r[]={'G','A','T','E','C','S'};
char *p,*q,*str1,*str2;
p=&s[4];
q=&r[2];
str1=r;
str2=s;
printf("%d",++*p-++*str1++ + ++*str2++-32);
printf("%c",3[s]+++*(s+2)-2[r]+++*q-++*r-32);
return 0;
}

 

Answer 1

Short Answer: 100q

Explaination:

1. Pointer 'p' points to address of s[4] = *(s+4) = *(4+s) = 4[s] {This relation between arrays and pointer is important}.

2. Pointer 'q' points to r[2].

3. Pointer str1 points to r[0]

4. Pointer str2 points to s[0]

 

So now we evaluate simply:

1. ++*p = ++(s[4]) = ++('c') = 'd' = 100.

2. ++*str1++ = ++(r[0])++ = ++('G')++ =' H'++ [ ASCII numerical value of 'H' will be taken in this step because pre-increment operator is used which first assigns the values and then increments ] = 72. Notice that after this step, 'H' would be turned to 'I' for next printf calculation.

3. ++*str2++ = 104 [and the array value changes to 'i' after assignment].

So, 100 - 72 + 104 -32 = 100. Since format specifier is %d, hence 100 will be printed.

 

4. For next printf, new array values are:

s[ ] = { 'i', 'a', 't', 'e', 'd', 's' }

r[ ] = { 'I', 'A', 'T', 'E', 'C', 'S' }

Now as mentioned earlier; 3[s] = *(3+s) = *(s+3) = s[3]. So applying the concepts, we get 113 after calculation. This for the case of %c results in q.

Since there is no newline charactor, output will be 100q

Comments

I have a doubt. In the second part of the print statement I'm getting

101+116-84+85-74-32=112=>p

Where am I going wrong ?

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++ [postfix] has more precedence than + [binary]. So -2[r]++ will be calculated first, then *q will be calculated; then plus. - (2[r]++) + (*q) is the real evaluation. This gives -84+84 = 0.

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But then, result is turning out to be 111.

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My bad. This is how it is being computed.

 

101+117-84+85-74-32.