Pointers

Question

Consider the declaration below:

typedef char *Str_typed;
#define Str_defined char*
Str_typed s1, s2;
Str_defined s3, s4;

Which one of the following statements is correct?
I. $s1$, $s2$, $s3$ and $s4$ are character pointers
II. $s1$, $s2$, $s3$ and $s4$ are characters
III. $s1$, $s2$, $s3$ are character pointers while $s4$ is a character
IV. None of the above

Comments

is it I. s1, s2, s3 and s4 are character pointers ??

---

It must be the first option. Use of typedef is well known. And since macros are dealt in pre-processing phase so even s3 and s4 are valid. Read typedef vs #define at the end of the link:

https://www.tutorialspoint.com/cprogramming/c_typedef.htm

Answer 1

answer is option 3.

typedef renames the data type of the variable . so str_typed will act as char pointer .(s1,s2 both of them are char pointers).

str_defined macro template will be replaced by its macro expansion during preprocessing .

str_defined will be replaced  by char*.

str_defined s3,s4; ===> char* s3,s4;

s3 is pointer whereas s4 is normal character variable. 

https://onlinegdb.com/rkmTwrG9G

Answer 2

Str_typed s1, s2;
Str_defined s3, s4;

that means s1 is of Str_typed

s2 is also Str_typed

hence s1 and s2 are pointer

while in #define 

char *s3 and char s4

so (C)