Why the functions doesn't return new line with replaceable keywords?

Question

Hey just doing some exercises in c, one is saying to replace tabs in the input string with any other characters , i restrict myself to only using getchar(), no gets() fgets() etc..., as my learning book didn't catch it yet, so i tried to not break the flow, the code below just printf() the same line it receives, can you please examine why ?

#include <stdio.h>



int main(){

        char line[20];
        char c;
        int i = 0;
        printf("Enter name: ");
        while ( c != '\n'){
                c = getchar();
                line[i] = c;
                ++i;}
        while (line[i] != '\0')
                if (line[i] == '\t')
                        line[i] = '*';

        printf("Line is %s \n", line);

        return 0;}

Your second `while` loop doesn't do anything with `i` - it just uses whatever value the previous `while` loop set it to last. — 500 - Internal Server Error, Jun 15 '21 at 16:31
The second `while` loop is reading array elements that have not been initialised or written to, and `i` may not even be within the array bounds. — Weather Vane, Jun 15 '21 at 16:33
You must also write the NUL termninator the second loop is looking for. — Weather Vane, Jun 15 '21 at 16:34
Note that [`gets()` is too dangerous to be used — ever!](https://stackoverflow.com/q/1694036/15168) so you should not bother to learn it (only learn that you should _never_ use it). — Jonathan Leffler, Jun 15 '21 at 16:56

score 1 · Accepted Answer · answered Jun 15 '21 at 16:33

1

c, which is used in c != '\n', is not initialized at first. Its initial value is indeterminate and using is value without initializng invokes undefined behavior.
You are checking line[i] != '\0', but you never assigned '\0' to line unless '\0' is read from the stream.
You should initialize i before the second loop and update i during the second loop.
Return values of getchar() should be assigned to int to distinguish between EOF and an valid character.
You should perform index check not to cause buffer overrun.

Fixed code:

#include <stdio.h>

#define BUFFER_SIZE 20

int main(){

        char line[BUFFER_SIZE];
        int c = 0;
        int i = 0;
        printf("Enter name: ");
        while ( i + 1 < BUFFER_SIZE && c != '\n'){
                c = getchar();
                if (c == EOF) break;
                line[i] = c;
                ++i;
        }
        line[i] = '\0';
        i = 0;
        while (line[i] != '\0'){
                if (line[i] == '\t')
                        line[i] = '*';
                ++i;
        }

        printf("Line is %s \n", line);

        return 0;
}

answered Jun 15 '21 at 16:33

MikeCAT

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Hey @MikeCAT thanks for the valuable explanations , i literately wrote it in my notebook , i still have 2 questions though: 1) why ```i+1 – Question Jun 15 '21 at 16:55
the lessons were amazing but your code still give the same input line indeed. – Question Jun 15 '21 at 17:04
@RidaEN-Nasry [This code works for me](https://wandbox.org/permlink/BGDuUhs4PAV8JoQN). – MikeCAT Jun 15 '21 at 17:06
@RidaEN-Nasry 1) `i+1 – MikeCAT Jun 15 '21 at 17:10
thanks a lot @MikeCAT one last question, when we assigned ```line[i] = '\0' ```, supposedly it was added as the last element to ```line[i]```? how's that ? i mean do array[i] or the i-th element of any array always refer to the last one? – Question Jun 15 '21 at 21:02
@RidaEN-Nasry `line[i] = '\0'` is assigning `'\0'` as the last character of the string because `i` is updated before that to make `line[i]` represent one element after the last assigned element. Generally `array[i]` is just an i-th element of `array` and there are no guarantee that it is the last element. – MikeCAT Jun 15 '21 at 22:55

Why the functions doesn't return new line with replaceable keywords?

1 Answers1