Thread: Char array

Code:

int main(void)
{
   .
   .
   .
   char row[15];
}

int function(char *row)
{
    char numbers[16] = { "0", "1", ........"15" };
   

   for(i=0; i<size; i++)
   { 
      if(....= ...)
           [Code help needed here! ]
  }
}

ok..i pasted a very small portion of my assignment that i'm doing...anywayz....the problem i'm have is trying to assign the "row" to a value from "numbers"....how wud i approach doing this, i tried several ways, they work, but from the numbers 1-9, any higher, it'll juz take the first number, i.e 15, will become 1.

this was one of them -> *row = *(numbers[i]); but yea, that's buggy...

but digits from 10 onwards are two, so i have to use " "....

i'm pretty sure, i can do " "...

no u cant, unless u use a pointer.
did u try compiling char numbers[15] ={"1","2"....

u will get an error.

anyways ,just use char numbers[15]={1,2,3,4...15}

this will take in both numbers.

Code:

Code:

char numbers[16];

That declares an array of 16 chars. A char can only hold one character. "15" is a 3 char array consisting of the chars '1', '5', and '\0'. It cannot fit in a char.

Originally Posted by qqqqxxxx

no u cant, unless u use a pointer.
did u try compiling char numbers[15] ={"1","2"....

u will get an error.

anyways ,just use char numbers[15]={1,2,3,4...15}

this will take in both numbers.

Code:

sorri, a typo! char* numbers[....]...hehe...anywayz how wud i fix it guys, thanks alot...

hmmm...well i only need to store only one of from 1-15...to row..so why do i need such a big array, actually cud u explain a bit about the double matrix....so it wudn't work with my way?

try this

Code:

char row[14];
row[12]= 15;
printf("%d",row[12]);

here is the output:
15

lol..my bad, i was assiging it to the wrong array, lol..thanks alot for the help, really appreciate the quick replies! ...