CODEWITHSUNDEEP
carraysstring
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Reputation: 4178

String arrays in C

I have an array of strings which when I iterate through and print its elements gives me unexpected results.

char currencies[][3] = {"EUR", "GBP", "USD", "JPY", "CNY"};

void show_currencies()
{
    int i;
    for(i=0; i<5; i++)
    {
        printf("%s - ", currencies[i]);
    }
}

when I call show_currencies() I get this on output.

EURGBPUSDJPYCNY - GBPUSDJPYCNY - USDJPYCNY - JPYCNY - CNY -

Can anyone explain this behaviour.

Thank you

Upvotes: 11

Views: 461

Answers (6)

Charles Beattie
Charles Beattie

Reputation: 5949

You are missing the nul terminators the strings are actually 4 characters long. Each string is then over writing the previous string's null terminator*. Try instead:

char currencies[][4] = {"EUR", "GBP", "USD", "JPY", "CNY"};

*As pointed out by caf it is not "over writing the previous string's null terminator" as the null terminator is never copied into the array. It is a fluke that the string is does not have garbled output after the final '-'.

Upvotes: 14

Matthew Flaschen
Matthew Flaschen

Reputation: 284786

You're declaring it wrong. This will work. It just lets the compiler set up an array of pointers-to-const-chars:

const char *currencies[] = {"EUR", "GBP", "USD", "JPY", "CNY"};

EDIT: Making it a two-dimension array, like Charles Beattie's answer, works too, provided you allocate space for the null. Also, specify that chars are const, per Christoph.

Upvotes: 8

codaddict
codaddict

Reputation: 454922

Change

char currencies[][3]

to

char currencies[][4]

strings in C are NULL terminated, to make their handling (in printing, copying etc) easier. example: char str[] = "ABC"; will declare a string of 4 char with \0 as the last char (index 3).

As a tip whenever on printing a char array you get unexpected results you might wanna check to see if the char array is NULL terminated or not.

Upvotes: 2

Henk Holterman
Henk Holterman

Reputation: 273169

You don't have an array of strings but an array of array-of-char. You could use:

char* currencies[] = {"EUR", "GBP", "USD", "JPY", "CNY"};  // untested

to allow for strings of different lengths.

Upvotes: 2

OscarRyz
OscarRyz

Reputation: 199195

My C is quite rusty, but try: 

char currencies[][3] = {"EUR\0", "GBP\0", "USD\0", "JPY\0", "CNY\0"};

I'm just curious to know what happens

Upvotes: 0

David Thornley
David Thornley

Reputation: 57036

Sure. "EUR" is four characters long - three for the letters, one for the terminating null character. Since you're explicitly specifying three-character arrays, the compiler is truncating, and so your data is strung together. You're lucky there is apparently a zero character at the end of the array, or you could get all sorts of garbage. Change your declaration to char currencies[][4].

Upvotes: 0

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