compile c++ code in c mode

Question

Following is my code saved as .cpp file and .c file

in .c it compiled fine, but threw the following error in .cpp

test.cpp:6: error: initializer-string for array of chars is too long
test.cpp:6: error: initializer-string for array of chars is too long

 

#include< stdio.h>

int main()
{

char str[2][2]= { "12", "12"};
int i;

for(i=0; i<2; i++)
printf("%d %s\n", i, str[i]);

return 0;
}

Is there any compiler directive or anything so that the c++ compiler takes this as C code itself.

I tried, extern "C", which didn't help.

Answer 1

Although it won't help your problem, you can select language to compile. With gcc it's -x flag, that needs to be followed by language. Like gcc -x c something.cpp ... will use c compiler to compile.

Answer 2

The character string "12" hold 3 places in C++ (In C too, BTW). You need one more char for the terminating '\0'.

char str[2][3]= { "12", "12"};

Answer 3

This would 'fit'

char str[2][2] = {
  { '1', '2' },
  { '1', '2' }
};

But you want this: https://ideone.com/wZB8F

char str[2][3]= { "12", "12"};

Otherwise, there is no room for the terminating null character

Equivalent:

char str[2][3]= { { '1', '2', '\0' },
                  { '1', '2', '\0' } };

Answer 4

C++ is a stricter language than C. The problem is that you create an array containing of arrays of two characters, then assign each sub-array three characters (the sequence '1', '2' and the string terminator '\0').

The array should be declared like this:

char str[2][3]= { "12", "12"};

The C compiler doesn't flag for this, and skips the string-terminator, so your printf statement will most likely print garbage after the string.

Answer 5

#include< stdio.h>

int main() 
{
    char str[][3]= { "12","12"}; 
    int i;

    for(i=0; i<2; i++) 
    {
        printf("%d %s\n", i, str[i]);
    }

    return 0; 
}

is the version that will work in C++....