Hi All,
Here is the code which generates Segmentation Fault. Can anyone explain why the third printf fails and the first printf works? - main()
-
{
-
char ch[7]={"Hello"};
-
char *p;
-
p=ch;
-
printf("Character is %c\n", *p);
-
printf("Character is %s\n", p);
-
printf("Character is %s\n", *p);
-
}
Output:
Character is H
Character is Hello
Segmentation Fault(coredump)
But Why *p could not fetch the entire string when I say %s in the third printf statement? Why same *p works for first printf?
Please explain...
Thanks & Regards
Sathish Kumar
3 1460 Ganon11 3,652
Recognized Expert Specialist
When you use the %s output modifier, I assume printf expects a char* (as that would point to a series of characters, or a string). But you passed it *p, which is a character. A character is not a string, and so the printf() statement fails.
Also, why don't you have a terminating \0 character at the end of ch? I'm surprised your second printf() statement works.
Banfa 9,065
Recognized Expert Moderator Expert
Also, why don't you have a terminating \0 character at the end of ch? I'm surprised your second printf() statement works.
Ganon you banana, have you got so into C++ you have forgotten C string syntax? "Hello" implicitly includes the '\0' character at the end of it and since the array it is being assigned to is long enough for the 6 characters in that string (it is 7 characters long) it wont be a problem.
Ganon11 3,652
Recognized Expert Specialist
Ganon you banana, have you got so into C++ you have forgotten C string syntax? "Hello" implicitly includes the '\0' character at the end of it and since the array it is being assigned to is long enough for the 6 characters in that string (it is 7 characters long) it wont be a problem.
Impossible! I can't forget something I never learned!
(never learned C except from TSDN :D)
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