C++ array not outputting full values

Question

So i'm experimenting with arrays when I come across a bit of a problem

code:

#include <iostream>
#include <windows.h>

using namespace std;

int main(int argc, char** argv) {
char items[] = {'bread', 'water', 'crisps', 'sweets', 'vegetables'};
for (int i = 0; i < strlen(items); i++) {
        cout << items[i] << endl;
    }
return 0;
}

What's happening is that when the code is ran, it's only outputting the last letter of each item, so 'd' 'r' 's' 's' 's'. I know i'm clearly doing something wrong here but I can't figure out what. I've been surfing on stackoverflow/google for a question like this but clearly what I have done is so obviously wrong, no one has asked!

Any help or a nudge in the right direction to particular documentation would be appreciated!

Thanks.

Use your compiler warnings (and errors).

Qaz
– Qaz

2014-05-04 16:51:39 +00:00
Commented May 4, 2014 at 16:51 — Qaz
– Qaz, Commented May 4, 2014 at 16:51
Also, en.cppreference.com/w/c/string/byte/strlen

juanchopanza
– juanchopanza

2014-05-04 16:53:11 +00:00
Commented May 4, 2014 at 16:53 — juanchopanza
– juanchopanza, Commented May 4, 2014 at 16:53
Are you aware of the difference between ' and "?

Kerrek SB
– Kerrek SB

2014-05-04 16:53:54 +00:00
Commented May 4, 2014 at 16:53 — Kerrek SB
– Kerrek SB, Commented May 4, 2014 at 16:53

Goz · Accepted Answer · 2014-05-04 17:00:21Z

1

Need an array of character pointers.
Need to use double quotes
Read a book on C++

i.e. code should be

#include <iostream>
#include <windows.h>

using namespace std;

int main(int argc, char** argv) {
char *items[] = {"bread", "water", "crisps", "sweets", "vegetables"};
for (int i = 0; i < (sizeof(items) / sizeof(*items)); i++) {
        cout << items[i] << endl;
    }
return 0;
}

edited May 4, 2014 at 17:00

Goz

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answered May 4, 2014 at 16:55

Ed Heal

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6 Comments

juanchopanza Over a year ago

const char* items[] would be better.

kec Over a year ago

sizeof(items) is a bug.

Andrew Matthew Over a year ago

You see, I've been looking at numerous things and some people were using double quotes and single quotes, so i just went with single (impatient i guess). I've just started learning C++ so i'm still trying to grasp the basics, thanks for the help all and sorry for it being such a naff and easy question!

Andrew Matthew Over a year ago

Also i'm just wondering why (sizeof(items) / sizeof(*items)) works? i understand the pointers part now but i'm still a bit stuck on sizeof, if someone could phrase it in simpleton terms, that'd be great! thanks

Blastfurnace Over a year ago

@AndrewMatthew: sizeof(items) is the size of the entire array or the size of five pointers to char. sizeof(*items) is the size of the first element of the array. So you have total size divided by the size of a single element to get the number of elements in the array.

|

Vlad from Moscow · Accepted Answer · 2014-05-04 17:04:18Z

0

Any of these items in the initializer list

{'bread', 'water', 'crisps', 'sweets', 'vegetables'};

is a multicharacter literal. According to the C++ Standard

A multicharacter literal, or an ordinary character literal containing a single c-char not representable in the execution character set, is conditionally-supported, has type int, and has an implementation-defined value.

You need to use string literals. I think that what you want is the following

#include <iostream>

int main(int argc, char** argv) 
{
   const char *items[] = { "bread", "water", "crisps", "sweets", "vegetables" };

   for ( const char *s : items ) std::cout << s << std::endl;

   return 0;
}

answered May 4, 2014 at 17:04

Vlad from Moscow

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