Print Char Array in C - only last letter?

Question

int main (){

    char array[2] = {'hola', 'adios'};  
    int i = 0;

    while (i<3){
        printf("%c", array[i]);
        i++;
    }

    return 0;
}

I don't why the output is the final letter of each word, like this: as:)

And it appears a smiley face, wtf?

I simply want to output hola adios

thats not a char array, that's a string array where u need to enclose `hola` in double quotes ( "hola" ) — exexzian, Dec 20 '13 at 13:57
if i use double quotes: excess elements in char array initializer — frankie3, Dec 20 '13 at 13:59
u need to change declaration part as well like this `char* array[2] ={ "hola", "adios"};` — exexzian, Dec 20 '13 at 14:00
Enable all warnings in your compiler (e.g. compile with `gcc -Wall -g`). It would have warned you.... — Basile Starynkevitch, Dec 20 '13 at 14:00

score 4 · Accepted Answer · edited May 23 '17 at 12:11

4

You need strings, not chars. hola is a string, not a char. Strings are surrounded in "" not in ''. So you need

const char* array[2] = { "hola", "adios" };

Now, this array has 2 elements, so loop through them

while (i<2){ /* also NOTE: 2 elements in the array */
    printf("%s", array[i]); /* note the "%s", it's not "%c" */
    i++;
}

I'd use for loop instead.

Why you don't have a compile-time error? See What do single quotes do in C++ when used on multiple characters? - it's similar in C.

edited May 23 '17 at 12:11

Community

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answered Dec 20 '13 at 13:58

Kiril Kirov

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+1 Had to think twice about the need for `const`, which I am certain is best. – chux - Reinstate Monica Dec 20 '13 at 15:45

score 2 · Answer 2 · answered Dec 20 '13 at 13:58

2

You're creating array of 2 single char elements. Declare array of two strings, and use double-quotes to declare strings, and not chars:

char* array[2] = {"hola", "adios"};

Also, in printf, use %s for printing out strings. And i should be in range {0, 1} - don't include 2.

answered Dec 20 '13 at 13:58

Nemanja Boric

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It's not invalid syntax, it's a multicharacter literal. – chris Dec 20 '13 at 14:00
Oh - didn't know that! Thanks! – Nemanja Boric Dec 20 '13 at 14:00
see http://stackoverflow.com/questions/7459939/what-do-single-quotes-do-in-c-when-used-on-multiple-characters – Kiril Kirov Dec 20 '13 at 14:00
@NemanjaBoric - I've reverted my down-vote, as I'm not 100% sure it's valid in C (I didn't mention it's not about C++). – Kiril Kirov Dec 20 '13 at 14:03
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@KirilKirov - don't worry - I didn't know if it is valid in the C++ - nice thing to learn today! – Nemanja Boric Dec 20 '13 at 14:04
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@KirilKirov, It's still there in C, it's just grouped together with a single character one as an *integer character constant*, as opposed to actually being named a *multicharacter literal* in C++. – chris Dec 20 '13 at 14:07

score 1 · Answer 3 · answered Dec 20 '13 at 13:58

1

char array[2] = {'hola', 'adios'};

This is not valid. You are in the Undefined Behavior zone.

You need a char** or char[] [] and use "" instead of '' for your strings.

answered Dec 20 '13 at 13:58

Arnaud Denoyelle

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What's undefined here? See http://stackoverflow.com/questions/7459939/what-do-single-quotes-do-in-c-when-used-on-multiple-characters – Kiril Kirov Dec 20 '13 at 14:00
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@KirilKirov Character literals are valid in C++. This question is about C. – Arnaud Denoyelle Dec 20 '13 at 14:01
Point taken, I've reverted my down-vote, as I'm not 100% sure it's valid in C. – Kiril Kirov Dec 20 '13 at 14:02
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@ArnaudDenoyelle, From the C99 standard: *An integer character constant is a sequence of one or more multibyte characters enclosed in single-quotes*. Anyway, I think the implementation-defined value of the constant being converted to a `char` could potentially be UB that would most likely just wrap. – chris Dec 20 '13 at 14:05

Print Char Array in C - only last letter?

3 Answers3