C. for loop with chars - c

Question

Can you explain these outputs?
1)
char s[]="TvNnFs",*p;
for(p=&s[5];p>=s;p--)
--*p;
puts(s);
OUTPUT: SuMmEr
2)
char s[]="TvNnFs",*p;
for(p=&s[5]; p>=s; p--)
((--*p)<'a') ? (*p+=('a'-'A')) : (*p);
puts(s);
OUTPUT:summer

Answer 1

This
char s[]="TvNnFs",*p;
where s is an array of characters and p is character pointer, is looks like below
s[0] s[1] s[2] s[3] s[4] s[5] s[6]
------------------------------------------
| T | v | N | n | F | s | \0 |
------------------------------------------
s 0x100 0x101 0x102 0x103 0x104 0x105 0x106.. (assume 0x100 is base address of s)
Next, the for loop
for(p=&s[5];p>=s;p--)
--*p;
Here p=&s[5] the pointer p points to address of s[5] i.e 0x105. Next is p>=s i.e 0x105 >= 0x100 which is true, then --*p executes i.e first *p that means value at 0x105 memory location which is s and decrements on that will makes s[5] as r.
Now char array s looks like
s[0] s[1] s[2] s[3] s[4] s[5] s[6]
---------------------------------------------
| T | v | N | n | F | r(new)| \0 |
---------------------------------------------
s 0x100 0x101 0x102 0x103 0x104 0x105 0x106..
Note : you might interested to know that by doing --*p how does s got affected ? Its because p is pointing or holding the address of s i.e whatever changes are done on *p will affect indirectly to s.
After that p-- happens i.e p now points to one location previous than before i.e 0x104. Same operation will happens until p reach to s i.e 0x100 >= 0x100. Finally char array s looks like
s[0] s[1] s[2] s[3] s[4] s[5] s[6]
-------------------------------------------
| S | u | M | m | E | r | \0 |
-------------------------------------------
s 0x100 0x101 0x102 0x103 0x104 0x105 0x106..
Hence it prints SuMmEr.
Case 2 :
char s[]="TvNnFs",*p;
for(p=&s[5]; p>=s; p--)
((--*p)<'a')?(*p+=('a'-'A')):(*p);
puts(s);
Here
s[0] s[1] s[2] s[3] s[4] s[5] s[6]
------------------------------------------
| T | v | N | n | F | s | \0 |
------------------------------------------
s 0x100 0x101 0x102 0x103 0x104 0x105 0x106..
|
p points here
For 0x105 >= 0x100 : this
((--*p)<'a')?(*p+=('a'-'A')):(*p);
is ternary operator i.e first ((--*p)<'a') executes, if it results in true then (*p+=('a'-'A')) will be the output else (*p). So Here it looks like ((--*p)<'a') i.e 'r' < 'a' which is false so just (*p) but s[5] got changed after this due to --*p.
s[0] s[1] s[2] s[3] s[4] s[5] s[6]
------------------------------------------
| T | v | N | n | F | r | \0 |
------------------------------------------
s 0x100 0x101 0x102 0x103 0x104 0x105 0x106..
|
p points here due to p--
Next, For 0x104 >= 0x100 : this
((--*p)<'a')?(*p+=('a'-'A')):(*p);
'E' < 'a' i.e 70 < 97 which is true so this (*p+=('a'-'A')) gets execute i.e
*p = *p + ('a' - 'A')
= 'E' + (97 - 65)
= 'E' + 32
*p = 'e' /* now s[4] overwritten by e(previously F) */
Now array looks like
s[0] s[1] s[2] s[3] s[4] s[5] s[6]
------------------------------------------
| T | v | N | n | e | r | \0 |
------------------------------------------
s 0x100 0x101 0x102 0x103 0x104 0x105 0x106..
|
p points here due to p--
Same operation hoes on until 0x100 >= 0x100.