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I have an array called array, and I want to copy a word that I have read in from some file and store it into that array at a certain index, how can I don it?

I thought when a file is read all input from the file is in "string" form. So if a number 100 is in the file its represented as "100".

I have tried the following:

#define maxlength 10

char array[10];
int i = 0;
char word[maxlength];

strcpy(array[i], word); //error
array[i] = word; //error

My guess is because the word is an array of characters it's not actually a string to copy through. Is there a way I can copy the word[] into the array[] in a string format? I.E. index 0 in array = ["hello"] index 1 = ["world"] etc.

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  • Post true code - a minimal reproducible example. What is maxlength? Commented Sep 2, 2019 at 3:49
  • 1
    "Is there a way I can copy the word[] into the array[] in a string format. i.e index 0 in array = ["hello"] index 1 = ["world"] etc." no. You've got an array of characters. The question is: how do you intend to pack hello in one character? Commented Sep 2, 2019 at 3:51
  • Have you tried strcpy(array, word); ? Commented Sep 2, 2019 at 4:21
  • char array[10] declares an array of 10-char. When referencing an element of an array [ ] acts as a dereference. So while array may be type char * (due to C11 Standard - 6.3.2.1(p3)), array[i] is type char resulting in your attempt to strcpy failing due to array[i] being an incompatible type. Commented Sep 2, 2019 at 4:40

2 Answers 2

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Your understanding of the concept of a "string" might be wrong.

A string in C is nothing else than an array of characters, with the NUL terminator (character '\0') at the end.

What you have here:

char array[10];

Is not an array of strings, it's just an array of characters. If you want to store an entire string inside array[0], then you'll have to declare your array as an array of arrays of char, like this:

char array[10][10];

Now, each element of your array can hold a string up to 9 characters long (remember, the last one has to be the terminator '\0').

To copy your word, you first have to be sure that it is a valid string (it should, again, end with a '\0'), and make sure it's less than 9 characters long (10 if counting the terminator), because you declared array[10][10]. Once you know that it's a correctly formed string, you can copy it:

strcpy(array[0], word);

If you don't know if word is correctly NUL-terminated, or you don't know if it's too long, then you can fisrt try to copy all the 10 characters with strncpy(), and then ensure it is terminated correctly putting a '\0' at the end manually:

strncpy(array[0], word, 10);
array[9] = '\0';
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1 Comment

Well, I did read the question - it was a bit vague, to be honest, and my answer was an attempt, at least! Still as it seems unpopular, I'll wipe it. But, some of your answer simply repeats what I already said. Hmmm.
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First let's clear some syntax errors then we can copy.

  1. cahr word[maxlength]; ---> char word[maxlength]; I hope you have defined maxlength = 10 somewhere because both length should be the same otherwise it will get run time exception.

  2. strcpy(array[i], word); //error ---> strcpy(array, word); strcpy expects destination and source both should be char* type and you are passing array[i] which is only char type as you access element of array by "[i]" So only use "array" in strcpy.

  3. array[i] = word //error ---> don't need this in array copy and even syntactically it is wrong you are trying assigning char* to char.

So our final code snippet would be like this:

char array[10];

char word[10] = {'T','h','a','n','k',' ','y','o','u'};

strcpy(array, word); // word is copied to the array

cout << array; ---> Thank you
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2 Comments

Your answer will crash (probably) as word is not null-terminated!
@Adrian as it happens there are 9 initializers for an array of size 10, and the rules are that any elements without initializer in this situation get set to 0 . However it's hard to understand why someone would write this instead of just using a string lteral

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