What is the correct way to convert an int array to string?

Question

I was thinking what the best way would be to convert a integer array, for example

int array[] = {0x53,0x74,0x61,0x63,0x6B,0x4F,0x76,0x65,0x72,0x66,0x6C,0x6F,0x77,0x00}

to a string, the above integer array is the equivalent of:

int array[] = {'S','t','a','c','k','O','v','e','r','f','l','o','w',0}

so a resulting string would be

std::string so("StackOverflow");

I was thinking of looping through eacht element with foreach and adding it to the string by casting it to char, but I'm wondering if there are cleaner / faster / neater ways to do this?

I think you meant char array[], not int array[].

barak manos
– barak manos

2014-02-18 19:08:45 +00:00
Commented Feb 18, 2014 at 19:08 — barak manos
– barak manos, Commented Feb 18, 2014 at 19:08
@barakmanos Why? Ints can store hex values too.

Paweł Stawarz
– Paweł Stawarz

2014-02-18 19:09:16 +00:00
Commented Feb 18, 2014 at 19:09 — Paweł Stawarz
– Paweł Stawarz, Commented Feb 18, 2014 at 19:09
Oh, sorry, didn't see the "equivalent" there...

barak manos
– barak manos

2014-02-18 19:09:50 +00:00
Commented Feb 18, 2014 at 19:09 — barak manos
– barak manos, Commented Feb 18, 2014 at 19:09

jrok · Accepted Answer · 2014-02-18 19:42:04Z

4

An int is implicitly convertible to char, you don't need any casts. All standard containers have constructors that take a pair of iterators, so you can pass the beginning and the end of the array:

std::string so( std::begin(array), std::end(array) );

This might not be any faster than a manual loop, but I think it fits the criteria of neater and cleaner.

There's a clean way to do the same with array of pointers, too. Use Boost Indirect Iterator:

#include <boost/iterator/indirect_iterator.hpp>

std::string s (
    boost::make_indirect_iterator(std::begin(array)),
    boost::make_indirect_iterator(std::end(array)) );

One more thing I just noticed - you don't need a 0 in the int array just to mark the end of a string. std::end will deduce the size of the array and 0 will actualy end up in the resulting string.

edited Feb 18, 2014 at 19:42

answered Feb 18, 2014 at 19:16

jrok

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1 Comment

UldisK Over a year ago

@Gizmo std::string so(array,array+length) should work - the end iterator is 1 past usable data, in this case array+length should point to 0x00 value in array;

barak manos · Accepted Answer · 2014-02-18 19:11:53Z

0

If you're looking for a (slightly) faster way, then you can do:

char str[sizeof(array)/sizeof(*array)];

for (int i=0; i<sizeof(array)/sizeof(*array); i++)
    str[i] = (char)array[i];

std::string so(str);

This will "save you" the repeated call to std::string::operator+=...

answered Feb 18, 2014 at 19:11

barak manos

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