Putting strings into a 2D chararray in C.

Question

How do I put strings into an 2D char array from (for example) a file?

char buffert[10][30];
int i = 0;

while(!feof(somefile)) {
  fscanf(somefile, "%s", temp);
  buffert[i][] = temp;
  i++;
}

This will not do it.

Answer 1

Try this:

while(i < 10 && fscanf(somefile, "%29s%*[^ \t\n]", buffer[i]) != EOF) {
  i++;
}

The "%29s" format specifier reads up to 29 characters into buffer[i], which is all it has space for. The "%*[^ \t\n]" reads and throws away non-whitespace, in case the string was longer than 29 characters.

Additionally, while (!feof()) { } is almost always the wrong thing to do, because EOF isn't set on the stream until end of file is encountered.

Answer 2

Stolen from this forum:

#include <stdio.h>

main()
{
  int dirsize = 80;
  char array[9][dirsize];
  int linenum = 0;
  FILE *filepointer;
  filepointer = fopen("projectfile.txt","r");
  while (feof(filepointer) == 0)
  {
    linenum++;
    fgets(array[linenum], dirsize, filepointer);
  }
  fclose(filepointer);
}

Another solution with a pointer to arrays instead of array of arrays is here.

Answer 3

Do away with the temp, and drop each string straight into the space you created for it (in buffert in this case).

char buffert[10][30];
int i = 0;
FILE * fp = fopen("myfile", "r");
while(!feof(fp)) { 
  fscanf(fp, "%s", buffert[i]);
  i++;
}

Here the file pointer is called fp. You will have to make a bunch of checks to prevent overflowing the 10 entries available in buffert, or the 30 chars available in each of its strings. You should also avoid feof().

Answer 4

it doesn't look to me like you're actually COPYING the string into an array. I'm surprised it compiles. am I missing something?

look into strcpy or strncpy, or you can copy using for loops.