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.

Write a program to produce the following output: 1 2 3 4 5 6 7 8 9 10

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0 Answers  

If the size of int data type is two bytes, what is the range of signed int data type?

0 Answers  

Do you know the use of 'auto' keyword?

0 Answers  

write a c program to remove all the duplicate characters in a string and replace with single character? ex:-input- AAABBBCCC output- ABC

2 Answers   HCL,