警告：格式％s期望为char *类型，但参数2为int类型

Question

I have already looked at other related questions, and none of them helped this case.
I am getting the warning listed in the title of my question, and my code for main is as follows:

int main( int argc, char *argv[] ) {

  char *rows;  
  int i, n;  

  printf("\nEnter the amount of rows in the telephone pad: ");  
  scanf("%d", &n);  

  rows = (char *) malloc( sizeof( char ) * n );  

  printf("\nNow enter the configuration for the pad:\n");  
  for( i = 0; i < n; i++ ) {  
    scanf("%s", &rows[i]);  
    printf("\n\t%s\n", rows[i]);  
  }  

  return 0;    
}

The user is to enter a number (say, 4), which will be scanned into n . The space is malloc'ed for the rows of the telephone pad. The user then will enter the n amount of rows for the configuration of the telephone pad. An example would be:

123
456
789
.0.

So I am confused as to why my last printf statement is getting this error.

Note: I also tried scanf("%s", rows[i]); : still got the error.
Note 2: I tried running the program anyways. Got a segmentation fault.
Note 3: I have #include <stdio.h> and #include <stdlib.h> at the top of my .c program.
Note 4: I have gcc'ed the program as such: gcc -ansi -pedantic -Wall tele.c .

Thank you for the help.

Answer 1

rows[i] isn't a char* -- it's not a "string".

(And you can't fit 3 characters (plus null terminator) in one character.)

Answer 2

As others have pointed out, you are allocating space for n chars, but you really want space for n rows with 4 chars each (3 characters entered by the user, and a null terminator).

There's a couple of ways to do this. You can first allocate n char * variables to point to the rows, then allocate 4 bytes for each row:

int main( int argc, char *argv[] )
{
  char **rows;
  int i, n;

  printf("\nEnter the amount of rows in the telephone pad: ");
  scanf("%d", &n);

  rows = malloc( n * sizeof rows[0] );

  printf("\nNow enter the configuration for the pad:\n");
  for( i = 0; i < n; i++ ) {
    rows[i] = malloc(4);
    scanf("%3s", rows[i]);
    printf("\n\t%s\n", rows[i]);
  }

  return 0;
}

Or, you can allocate n 4-character arrays up front:

int main( int argc, char *argv[] )
{
  char (*rows)[4];
  int i, n;

  printf("\nEnter the amount of rows in the telephone pad: ");
  scanf("%d", &n);

  rows = malloc( n * sizeof rows[0] );

  printf("\nNow enter the configuration for the pad:\n");
  for( i = 0; i < n; i++ ) {
    scanf("%3s", rows[i]);
    printf("\n\t%s\n", rows[i]);
  }

  return 0;
}

Answer 3

your printf is passing in a char, not a %s

you are saying get the i'th char in the string 'rows'.

More importantly, your whole technique is going to fail badly....

I think you want an array of strings.... or you want to change your scanf to a %c