Traverse through two dimensional array with variable sized strings in C

Question

What am I doing wrong here? The first one works like a charm the second one crashes.

char *t[3][4] = {
  {"one", "two", "three", "Four"},
  {"Five", "Six", "Seven", "Eight"},
  {"Nine", "Ten", "Eleven", "Twelve"}
};
int i,j;
for (i=0;i<3;i++) {
  for (j=0;j<4;j++) {
    printf("%s\n",t[i][j]);
  }
}

Same array just want to use in function:

foo(t, 3, 4);

char **foo(char **t, int row, int col) {
  int i,j;
  for (i=0;i<row;i++) {
    for (j=0;j<col;j++) {
      printf("%s\n",t[i][j]);
    }
  }
}

Process terminated with status -1073741510 (0 minute(s), 8 second(s));

char **foo(char **t, int row, int col) --> void foo(int row, int col, char *t[row][col]) (call foo(3, 4, t);) — BLUEPIXY
– BLUEPIXY, Commented Jul 3, 2016 at 2:49
@BLUEPIXY: You don't even have to specify 'row' there since C's row major indexing will automatically take care of the first dimension, if the second is passed to it. — cs95
– cs95, Commented Jul 3, 2016 at 3:12
I'm not sure if you meant to pass char *** (which would be incorrect as a 2D array of char * is not a pointer-to-pointer-to-pointer-to-char) or just thought a char ** would refer to an array of char * regardless of the number of dimensions (also not right, but a different misunderstanding)... — Dmitri
– Dmitri, Commented Jul 3, 2016 at 3:21
Did the compiler not warn you about type mismatches? It should have done. Fundamentally, the type of t in the argument list of foo does not match the type of t, the array. When I compiled your code, I got errors (because I convert warnings to errors with -Werror) like: asp19.c:22:9: error: passing argument 1 of ‘foo’ from incompatible pointer type [-Werror=incompatible-pointer-types] identifying the line foo(t, 3, 4); and observing asp19.c:3:6: note: expected ‘char **’ but argument is of type ‘char * (*)[4]’ referring to void foo(char **t, int row, int col); — Jonathan Leffler
– Jonathan Leffler, Commented Jul 3, 2016 at 4:19

cs95 · Accepted Answer · 2016-07-03 03:10:20Z

2

Dealing with pointers is cumbersome, especially with string matrices. It's best to write your function in such a way that the compiler clearly knows what's being passed to it.

Change your function signature to void foo(int row, int col, char* t[][col]) { ... }, and correspondingly, call your function as foo(3, 4, t);. You're clearly telling the compiler what's being passed.

answered Jul 3, 2016 at 3:10

cs95

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Comments

Attila Szkaza · Accepted Answer · 2016-07-03 10:59:42Z

0

This is my code witch worked with char **foo(char **t, int row, int col). The problem was that i had a ** array with a pointer to its beginning, and you cant traverse it like a simple 2d array like t[i][j]. But you can use i*col to reach the row you want, and the j is the locating the j-th element in i-th row. Extra note that the size of the containing strings doesnt matter because you have addresses in the array . Thanks for the helpful comments witch led me to the correct answer.

for (i=0;i<row;i++)
    {
        for (j=0;j<col;j++)
        {
        printf("%s",*(t+i*col+j)); 
        }
    }

answered Jul 3, 2016 at 10:59

Attila Szkaza

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