converting char 1d array to 2d array

In the loop, the elements in the 2d array, d has the right strings. but once it exits the loop, all the elements in d are the last string.. how should i resolve this?

Code:

---

#include<stdio.h>
#include<string.h>

int main (void)
{
char s[]="abcd:efgh:ijkl:\0";
char *d[100];
printf("s is %s\n",s);
char wasi[]="wasi";
char ahmed[]="ahmed";
int k = strlen(s); int x=0,y=0; int z=0;
char c = s[x]; char test[1000]; //memset(test,"0",sizeof(test));
while(x<k)
{
while(c!=':')
{
if(c=='\0')
{break;}
test[y]=c;
x++;y++;
c=s[x];
}
test[y]='\0';
//test contains a string between colons
printf("test is %s\n",test);
//printf("z is %d\n",z);

d[z]=test;
strcat(d[z],"\0");
printf("d[%d] is %s\n",z,d[z]);

z++;
x++;c=s[x];
y=0;
if(c=='\0')
{break;}
}
d[z]="\0";
// printing the contents of the string
int i;
for( i=0; i<2;i++)
{
printf("d[%d] = %s\n",i,d[i]);
}
// EVERYTHING INSIDE d IS THE SAME AS THE LAST STRING


return 0;


}

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Please use code tags when posting code!

You want the first row to be "abcd", second row to be "efgh", third to be "ijkl", in a char array?

You seem to have the algorithm basically set up. I'm not clear what the problem is that you're having.

Using the input in your program, what are getting for output, and what do you want for correct output?

Even with code tags, the indentation is dog-food.

Figures, it's also cross-posted Char 1d to 2d array problem C programming? - Yahoo! Answers

No doubt, the code here was a copy/paste from there to here, and in the process, and semblance of formatting that it might have had has been lost.

Here, read this before deciding to spam forums again.
How To Ask Questions The Smart Way

k this is the first time im using this..so take it easy..
The input is abcd:efgh:ijkl

expected output:
abcd
efgh
ijkl

output produced:
ijkl
ijkl
ijkl

why is this happening?

Quote:

---

Originally Posted by wasi.cjr

k this is the first time im using this..so take it easy..
The input is abcd:efgh:ijkl

expected output:
abcd
efgh
ijkl

output produced:
ijkl
ijkl
ijkl

why is this happening?

---

Because of the logic that is used -- not quite right.

First, where is the 2D char array? I don't see one. Or do you just want it printed up in rows, like you show for the output, above?

@Salem - thanks for the code tags, however. That's a lot better than before!

isnt char *d[100]; a 2-D array?.. i thought this is a 2d array and i had to convert a 1D array to 2D array where ':' is the token

Quote:

---

Originally Posted by wasi.cjr

isnt char *d[100]; a 2-D array?.. i thought this is a 2d array and i had to convert a 1D array to 2D array where ':' is the token

---

It is. I have a rough time with code not in code tags.


Let's redo the variables to something more meaningful. Row and col for their respective indices, c for char, and len for the total length of the original string.

Code:

---

#include <stdio.h>

int main(void) {

  int len;
  int i = 0;
  int row = 0;
  int col = 0;
  char s[] = {"abcd:efgh:ijkl"};
  char d[3][5];
  char c;
  printf("\n\n");
  len = strlen(s);
  while(i<len) {
    c = s[i];   
    if(c==':') {
      d[row][col]='\0';
      row++;
      col = 0;
    }
    else {
      d[row][col]=c;
      ++col;
    }
    ++i;
  }
  d[row][col]='\0';
  //d[row] contains a string between colons
  for(i=0;i<=row;i++)
    printf("d[%d] is %s\n",i, d[i]);

  (void) getchar();
  return 0;
}

---

char *d[5] is not a true 2D char array. It's (reading from right to left: a 5 element array of pointers to char.

That's very useful sometimes, but then each of those pointers needs to be:

1) given a valid address
and
2) given a number of memory addresses for their row of 5 chars.

which is not what I would want to do for a program like this.

Quote:

---

Originally Posted by Adak

It is. I have a rough time with code not in code tags.


Let's redo the variables to something more meaningful. Row and col for their respective indices, c for char, and len for the total length of the original string.

Code:

---

#include <stdio.h>

int main(void) {

  int len;
  int i = 0;
  int row = 0;
  int col = 0;
  char s[] = {"abcd:efgh:ijkl"};
  char d[3][5];
  char c;
  printf("\n\n");
  len = strlen(s);
  while(i<len) {
    c = s[i];   
    if(c==':') {
      d[row][col]='\0';
      row++;
      col = 0;
    }
    else {
      d[row][col]=c;
      ++col;
    }
    ++i;
  }
  d[row][col]='\0';
  //d[row] contains a string between colons
  for(i=0;i<=row;i++)
    printf("d[%d] is %s\n",i, d[i]);

  (void) getchar();
  return 0;
}

---

char *d[5] is not a true 2D char array. It's (reading from right to left: a 5 element array of pointers to char.

That's very useful sometimes, but then each of those pointers needs to be:

1) given a valid address
and
2) given a number of memory addresses for their row of 5 chars.

which is not what I would want to do for a program like this.

---

Thanks a lot!

You can of course just pretend it's a 2D array:

Code:

---

char foo[ X ];

foo[ (row * rowlen) + col ]

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This isn't want you are trying to do here, but you could.


Quzah.