question about C 'for' statement ...

purpleburple · 08-02-2002, 02:15 AM

Hi. How does an array get stored?

say
char array[ ] = "linux";

array[0] is 'l'
array[1] is 'i'
etc,

but is there a '\0' in the 5th cell or EOF
Question arises cause I got a segmentation fault in the following>

char array [ ] = "linux";

for( i = 0; i != '\0'; i++)
printf("array[%d] contains %c\n", i, array[i]);

the output just kept printing a long list of array contents to stdout and ended saying segmentation fault

I probably am not checking for EOF correctly in the 'for' loop.
Say char array[10] was to be filled in by the program asking user for input?
I would'nt want to print the contents of all the cells he or she didnt fill ( say if user only types name 'Bill' into the array) How ... using the 'for' loop would I check to stop 'i' from further augmenting at end of the array?
I thought it was 'i != '\0' ' or i != EOF

Can someone help?

thanks

Mara · 08-02-2002, 05:03 AM

Code:

#include <stdio.h>

int main() {
	char tab[]="linux";
	int i=0;
	for(i=0;tab[i]!='\0';i++)
		printf("tab[%d]: %c\n",i,tab[i]);
	return 0;
}

And it's output:

Code:

tab[0]: l
tab[1]: i
tab[2]: n
tab[3]: u
tab[4]: x

So, where's the problem?

sarin · 08-02-2002, 06:43 AM

Hi Mara, he is trying to compare i != '\0'. Thats where it fails.
--Sarin

purpleburple · 08-02-2002, 10:41 AM

thank you Mara. I was doing as sarin said. 'i' wouldn't equal '\0'. Forgive me?
Im new to C. But I love it!

thanks for your help.

Mara · 08-02-2002, 05:56 PM

I wrote that short program and was wondering why purpleburple has a problem. I haven't realized this 'i != '\0'. Don't worry, purpleburple, such mistakes are very common...

guest72485 · 08-14-2002, 03:05 AM

Quote:

Originally posted by Mara

Code:

char tab[]="linux";

how does a null terminating character gets introduced?

what would happen if i had the following declaration

char tab[5]="linux";

even in this case wil a null will be introduced? if yes then does that mean the size if incremented. if no then will the following code segment work?

for(int i=0;tab[i] != '\0';i++)
printf(" [%d] = [%c]",i,tab[i]);

tundra · 08-14-2002, 03:54 AM

char tab[] = "linux";
automatically introduces the '\0' since you did not initially specify the number of characters you wanted in the array.

but if you explicitly specify
char tab[5];
then you will only be allocated an array of 5 char sized "spaces". if you don't explicitly specify the '\0', then it will produce a segmentation fault. what happens is most likely, your program will try to insert the '\0' and end up writing over another spot in memory that may be in use(or not, depending). since this is quite dangerous, i think the program will be halted by the OS in the case of linux.

Mara · 08-14-2002, 04:04 AM

Quote:

Originally posted by guest72485

char tab[5]="linux";

Dangerous. '\0' will not be introducted, so using such a string will cause trouble.

abi_sh · 08-25-2002, 07:12 AM

hey y don't u try array[i]!='\0' its not with i which u r comparing its with the value with which u r checking the condition ,so try giving this in condition it will work