Answer / shruti

the syntax of malloc is wrong.

in your example it should be written as:

char *p;

p = (char *)malloc(sizeof(char));

Answer / guest

The pointer you declared is p, not *p.

Answer / parasher

char *p;

p = (char *)malloc(sizeof(char));

Answer / raj

here syntox of malloc is not fallowed

Answer / clay

Here,

After char *p;
since pointer p is not initialized it is pointing at some
unknown location.

In the next step, the address of the memory allocated by
malloc() is stored at some garbage location pointed by p.

Here p is never initialized or never assigned any value.

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