CODEWITHSUNDEEP
cpointers
Amit Gupta
Amit Gupta

Reputation: 287

What should be the output?

I'm expecting h here.But it is showing g. Why?

char *c="geek"; 
printf("%c",++*c);

Upvotes: 4

Views: 101

Answers (3)

NPE
NPE

Reputation: 500673

You are attempting to modify a string literal, which is undefined behaviour. This means that nothing definite can be said about what your program will print out, or indeed whether it will print anything out.

Try:

char c[]="geek"; 
printf("%c",++*c);

For further discussion, see the FAQ.

Upvotes: 9

Sankar Mani
Sankar Mani

Reputation: 178

The expression will be evaluated like this (++(*c)),

First the inner *C will be evaluated so it will print g. then the increment operator will be applied to the pointer variable C. After this print statement the pointer c will point to the next character 'e'.

Upvotes: -1

Omkant
Omkant

Reputation: 9214

It's undefined behaviour since you are trying to modify the string literal

*c will give character 'g' but when you apply this ++*c means you are trying to do

*c=*c+1; which is modifying the string and its undefined in language standard

It's better to use char array to solve this since "string" literal are stored in readonly memory and modifying it causes this.

Upvotes: 0

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