C Programming - Strings - Discussion

int main()
{
static char s[25] = "The cocaine man";
int i=0;
char ch;
ch = s[++I]; //pre-increment pointer from s[0] to s[1] i.e s[1]= h
printf("%c", ch); // print s[1]=h
ch = s[i++]; //post-increment pointer from s[1] to s[2]

printf("%c", ch);// print s[1]=h because ch = s[i++]; is post increment which increment after next print

ch = i++[s]; //post-increment pointer from s[2] to s[3]
printf("%c", ch); // print s[2]=e because ch = s[i++]; is post increment which increment after next print

ch = ++i[s];// this increment s[3] which is blank space by incrementing the ascii address 20 to 21(!)
printf("%c", ch);//print (!)
return 0;
}

//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.

Here for first ch=s[++i]: first i=1,then ch=s[1]='h'.
For second ch=s[i++]:first ch=s[i]=s[1]='h',then i++becomes i=2;.
For third ch=i++[s];ch=2[s]='e';then i++ becomes i=3;.
For fourth ch=++i[s];here first ch=3[s]=s[3]=' '(space), ASSCII value=32 after that ++i[s] Takes place so ch=33 which means ASSCII Value for '!' so ch='!'.

++i means that i is incremented first and then evaluated.

i=0;
if(++i == 1) this is evaluated as ((i+1) == 1) which is ((0+1) == 1) which is true.

i++ means that i is evaluated first and then incremented.

i=0;
if(i++ == 1) this is evaluated as (i == 1) which is (0 == 1) which is false and then i = i+1;

++i is pre incrementor.

Whereas i++ is postincrementor, both of these are the binary operator and has LOW PRIORITY when compared to Assignment operator(=),so in 1st case, ++i yields s[1], and then due to priority case first " VALUE IS ASSIGNED AND THEN INCREMENTED".

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[3] is the space in the string "The cocaine mam" i.e
s[0]==T
s[1]==h
s[2]==e
s[3]==(space) so the ascii value for space was 32,

Since ++i[s] == ++ (i[s]) == ++ (s[i])== ++ (s[3]), so ++(32)==33.

ASCII value 33 will print === !

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 !.

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).

Why it didn't take s[4]=c and gave the answer hhec directly, I mean to say as it did in the first 3 cases, why not it's the same as in the last case. Why do they take ASCII value in the last?