How to change characters within an array of strings?

Question

I'm quite new to C and as part of my task, I must remove the the newline character that is added at the end of a string using fgets().

My lecturer advised this method:

char names[20];
fgets(names,20,fp);
names[strlen(names)-1]='\0';

I've omitted some of the code as I'm only demonstrating the method.

However, in our task, we have to deal with arrays of strings. I've tried...

names[strlen(names[i])-1]='\0';

but seems to only find the length of the string, deduct 1 and then use that number in the index which then sets a string further down the array to \0

Is there a way to access the individual characters of the strings?

I know I can access the strings using

names[i] // where i is the numeric index

but I need to access the individual characters within that string.

This is my first time posting on StackOverflow so please inform me if I haven't included enough detail or have formatted my question poorly.

Thanks in advance.

Your teacher's code will not strictly remove a newline. See my comment in @erosenin's correct answer below if you are interested. — Darren Stone
– Darren Stone, Commented Dec 29, 2013 at 23:19

erosenin · Accepted Answer · 2013-12-29 20:02:57Z

4

As noted, in C a string is an array of characters. So, you can access the individual characters like you access any array. For your task what you need to do is

  char names[100][20];
  //Now this declares 100 strings of size 20 each.
  //To access a single character in the position '4' from string 0 you can write
  printf("%c",names[0][3]);

  //To modify the string in position 'i' you will use 
  names[i][strlen(names[i])-1]='\0';

The last line is very similar to what you have written for a single string.

With the first index 'i' you access the string in that position. and with the second index you access a particular character of that string.

answered Dec 29, 2013 at 20:02

erosenin

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4 Comments

Alluziion Over a year ago

I considered a 2D Array but wasn't sure which way to approach it in the question but thanks to your help, I attempted it and it's worked perfectly! Thank you.

erosenin Over a year ago

No problem! Glad, I could be of help.

Darren Stone Over a year ago

Solid answer. This is not a critique of your answer; more for OP. To treat the input data with a bit more care, I would suggest a strlen comparison or content check before assuming the len-1 position contains a newline. You can end up losing a letter of a 19-character name unnecessarily this way. Sure, it only truncates one letter in the worst case, but it can be prevented. Cheers.

Darren Stone Over a year ago

I should clarify. I'm referring to behaviour of the fgets function -- it does not guarantee a newline. OP's question was using fgets to fill the names in.

haccks · Accepted Answer · 2013-12-29 20:10:10Z

1

Try this:

#include <stdio.h>
#include <string.h>

int main(void)
{
    char names[10][10];
    char c = 'A';
    for(int index = 0; i < strlen(names) - 1)
    {
        names[0][index] = c++;
    }
    names[i][strlen(names[i])-1]='\0';

    printf("%s", namws[0]);

    return 0;
}

edited Dec 29, 2013 at 20:10

answered Dec 29, 2013 at 19:49

haccks

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4 Comments

user529758 Over a year ago

I'm sorry but this doesn't answer the question. See BLUEPIXY's comment on the question.

haccks Over a year ago

@H2CO3; I think he is talking about 1D array.

user529758 Over a year ago

No, he isn't. "However, in our task, we have to deal with arrays of strings. [...] but seems to only find the length of the string, deduct 1 and then use that number in the index which then sets a string further down the array to \0" - that's quite the description of the problem arising when one confuses an array of strings with an array of characters.

haccks Over a year ago

@H2CO3; Mention not :)

AbhinavRanjan · Accepted Answer · 2013-12-29 19:53:01Z

0

In c, string is actually an array of char. So,

char[] charArray = "string";

charArray[0] will give you 's'.

answered Dec 29, 2013 at 19:53

AbhinavRanjan

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