making a char array error

Question

in a larger code i have this code:

#define N 10
..
..
..
char Map [N][N] =              {"##########",
                                "#@       #",
                                "#        #",
                                "#        #",
                                "#        #",
                                "#        #",
                                "#        #",
                                "#        #",
                                "#        #",
                                "##########"};

when i try to compile and run it form Code bloke i gives me the error

F:\C++\Maze\main.cpp|25|warning: extended initializer lists only available with -std=c++11 or -std=gnu++11 [enabled by default]|

any idea what i did wrong in this simple code?

a warning is not an error (yet).

glglgl
– glglgl

2014-08-15 07:46:22 +00:00
Commented Aug 15, 2014 at 7:46 — glglgl
– glglgl, Commented Aug 15, 2014 at 7:46

Miguel Prz · Accepted Answer · 2014-08-15 07:20:38Z

6

The length of each row in your example is 10+1=11 (remember the null terminator for string needs space to allocate). So:

#define N 10
#define M 11
char Map [N][M] =              {"##########",
                                "#@       #",
                                "#        #",
                                "#        #",
                                "#        #",
                                "#        #",
                                "#        #",
                                "#        #",
                                "#        #",
                                "##########"};

answered Aug 15, 2014 at 7:20

Miguel Prz

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Comments

Bathsheba · Accepted Answer · 2014-08-15 07:22:29Z

4

You're forgetting that a string literal has an implicit null-terminator so requires an extra byte of storage.

If you had written char Map [N][N + 1] then it will work fine.

Even better, let the compiler count the rows: write char Map [][N + 1]. You might want to change the definition of N. N + 1 will be evaluated at compile time which is why you can use it as an array dimension.

answered Aug 15, 2014 at 7:22

Bathsheba

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Comments

aaragon · Accepted Answer · 2014-08-15 07:32:30Z

0

Why not using a vector of strings? You can still reference the elements inside by using operator[]:

#include <string>
#include <iostream>
#include <vector>

using std::cout;
using std::endl;

int main(){

std::vector<std::string > s =  {{"##########"},
  {"#@       #"},
  {"#        #"},
  {"#        #"},
  {"#        #"},
  {"#        #"},
  {"#        #"},
  {"#        #"},
  {"#        #"},
  {"##########"}};

  for (auto i : s)
      cout<<i<<endl;

  cout<<"\nGet @ symbol:"<<s[1][1]<<end
  return 0;
}

Outputs:

##########
#@       #
#        #
#        #
#        #
#        #
#        #
#        #
#        #
##########

Get @ symbol:@

You can find the code here

answered Aug 15, 2014 at 7:32

aaragon

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Comments

David C. Rankin · Accepted Answer · 2014-08-15 07:23:22Z

-1

You need to create Map as follows:

char *Map [N] =            {"##########",
                            "#@       #",
                            "#        #",
                            "#        #",
                            "#        #",
                            "#        #",
                            "#        #",
                            "#        #",
                            "#        #",
                            "##########"};

For example:

#include <stdio.h>

#define N 10

int
main (void) {

    int i;
    char *Map [N] =            {"##########",
                                "#@       #",
                                "#        #",
                                "#        #",
                                "#        #",
                                "#        #",
                                "#        #",
                                "#        #",
                                "#        #",
                                "##########"};

    for (i = 0; i < 10; i++)
        printf (" Map[%d]  %s\n", i, Map[i]);

    return 0;
}

output:

$ ./bin/tcs
 Map[0]  ##########
 Map[1]  #@       #
 Map[2]  #        #
 Map[3]  #        #
 Map[4]  #        #
 Map[5]  #        #
 Map[6]  #        #
 Map[7]  #        #
 Map[8]  #        #
 Map[9]  ##########

answered Aug 15, 2014 at 7:23

David C. Rankin

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5 Comments

Bathsheba Over a year ago

That's different: in this case the data block will be read-only memory and attempts to modify it is undefined behaviour. The way the OP has done it is fine.

David C. Rankin Over a year ago

Hmm, is there any reason you can't just make it constant with const char *Map [N] = ...?

David C. Rankin Over a year ago

Well I apologize, the subtleties of the question were missed, the first simple solution was to declare an array of pointers and fill them with the literals supplied allowing null terminator to be added without restriction. Why is that different from explicitly declaring the column size one bigger? (I"m curious as to how that would change the use/implementation?)

David C. Rankin Over a year ago

OK - I understand. *(Map []) + N + 1 does not equal *Map[] Sorry for the misread of the question.

Jim Balter Over a year ago

"is there any reason you can't just make it constant with const char *Map [N] = ...? " -- The reason is that the OP presumably wants to alter Map, which is just being initialized with its starting value.

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