C Programming - Strings - Discussion

Thanks for the explanation @Tarlika.

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?

++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".

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;
}

How hhe! came? Please explain me.

Nice explanation. Thank you all.

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='!'.

Why s[i]=i[s]?

++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;

Nice explanation @ Ragini.