C Programming - Strings - Discussion
Discussion Forum : Strings - Find Output of Program (Q.No. 13)
13.
What will be the output of the program ?
#include<stdio.h>
int main()
{
static char s[25] = "The cocaine man";
int i=0;
char ch;
ch = s[++i];
printf("%c", ch);
ch = s[i++];
printf("%c", ch);
ch = i++[s];
printf("%c", ch);
ch = ++i[s];
printf("%c", ch);
return 0;
}
Discussion:
34 comments Page 2 of 4.
Madhu said:
1 decade ago
The o/p is hhe! because ch=++i[s];
Firstly i is value is 3 from i++ then we use ++i[s]
++i is evaluated its space ascii value is incremented and it becomes !.
So the o/p is hhe!
Firstly i is value is 3 from i++ then we use ++i[s]
++i is evaluated its space ascii value is incremented and it becomes !.
So the o/p is hhe!
Iranna said:
1 decade ago
s[++i]=s[1]=h here i is incremented and assigned
s[i++]=s[1]=h here i is incremented but not assigned
i++[s]<->(i++)[s]<->s[i++]=s[2]=e
++i[s]<->++(i[s])<->++(s[i])=++(s[3])=++(asci value of space)=++(32)=33 is equivalent to !
So ans is hhe!
s[i++]=s[1]=h here i is incremented but not assigned
i++[s]<->(i++)[s]<->s[i++]=s[2]=e
++i[s]<->++(i[s])<->++(s[i])=++(s[3])=++(asci value of space)=++(32)=33 is equivalent to !
So ans is hhe!
Xyz said:
1 decade ago
Can anyone explain clearly.
Shashi said:
1 decade ago
[] has higher precedence than ++.
so i++[s] <---> (i++)[s] <---> s[i++]
++i[s] <---> ++(i[s]) <---> ++s[i]=33 whose ascii is !
so i++[s] <---> (i++)[s] <---> s[i++]
++i[s] <---> ++(i[s]) <---> ++s[i]=33 whose ascii is !
FlameNfire said:
1 decade ago
int i=0;
std::cout<<i++;
//Adds 1 to i, but output i's value before adding.
int a=0;
std::cout<<++a;
//Adds 1 to a, but output a's value after adding.(Which seems more logical for us).
std::cout<<i++;
//Adds 1 to i, but output i's value before adding.
int a=0;
std::cout<<++a;
//Adds 1 to a, but output a's value after adding.(Which seems more logical for us).
Preethi said:
1 decade ago
What is the actual explanation. Please explain it clearly for each line. What is ASCII value. How you are calculating it. Tell me clearly?
Ragini said:
1 decade ago
s[++i] = given i=0 ++i means 0+1 therefore s[1]=h.
s[i++] = s[1] itself because post increment therefore s[1]=h.
i++[s] = s[i+1] as now i=1 increment i => 1+1=2 s[2]=e.
++i[s] = ++(s[i]) => ++s[2] => 33(ascii value ) so !.
s[i++] = s[1] itself because post increment therefore s[1]=h.
i++[s] = s[i+1] as now i=1 increment i => 1+1=2 s[2]=e.
++i[s] = ++(s[i]) => ++s[2] => 33(ascii value ) so !.
Ravli said:
1 decade ago
I didn't understand it clearly can anyone explain it?
Manohar Reddy said:
1 decade ago
//Remember %c for number means it will print equivalent ASCII value//.
I=0; s[++i] ==> s[1] ==> h.
I=1; s[i++] ==> s[1] ==> h /*Since it is post increment no change to inside i*/.
I=2; i++[s] ==> (i++) [s] ==> s[i++] ==> s[2] ==>e /*post increment to I */.
I=3; ++i[s] ==> ++ (i[s]) ==> ++ (s[i]) ==> ++ (s[3]) ==> ++ (32) ==> 33.
==> ASCII value of 33 !
S[3] is nothing but "space" in the given string. ASCII value for "space" is 32.
It was incremented by 1. So 32+1=33.
ASCII value of 33 will be print.
I=0; s[++i] ==> s[1] ==> h.
I=1; s[i++] ==> s[1] ==> h /*Since it is post increment no change to inside i*/.
I=2; i++[s] ==> (i++) [s] ==> s[i++] ==> s[2] ==>e /*post increment to I */.
I=3; ++i[s] ==> ++ (i[s]) ==> ++ (s[i]) ==> ++ (s[3]) ==> ++ (32) ==> 33.
==> ASCII value of 33 !
S[3] is nothing but "space" in the given string. ASCII value for "space" is 32.
It was incremented by 1. So 32+1=33.
ASCII value of 33 will be print.
Anusha.v said:
1 decade ago
In the last statement ++i[s] is 33. I didn't understand explanation of this statement.
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