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Hello How to print string which is passed from 2d array to function called print_2d and in that function i am catching base address of 2d array which has string as its element

now with pointer to an array now i want to print the string using pointer to an array? my code is below

#include<stdio.h>
void print_2d(char (*p)[10],int r) // i am catching with pointer to an array
{
    int i;
    for(i=0;i<r;i++)
    printf("%s",*p[i]); //segmentation fault why?
                            // how to print then?
}

main()
{
    char *s[10]={"one","two","three","four"};
    int r;
    r=sizeof(s)/sizeof(s[0]);
    print_2d(s,r);
}
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4
  • Do you want a pointer to an array or an array of pointers? Commented Aug 16, 2014 at 3:40
  • @Code-Apprentice pointer to an array (*p)[10]? Commented Aug 16, 2014 at 4:47
  • 2
    But s is declared as an array of pointers... Commented Aug 16, 2014 at 4:55
  • Note that there is no 2D array present in the code shown. There is an array of pointers, but that's a 1D array. Commented Aug 17, 2014 at 7:17

4 Answers 4

Reset to default
2

Your print_2d function is declared as accepting a pointer to char[10] array. However, nowhere in your code there's a char[10] array. There's noting in your code that can be legally passed to print_2d.

Anyway, it is not possible to "fix" your code, until you explain what you are trying to do.

Judging by the body of print_2d and by what you have in main, you might need the following 

void print_2d(char *(*p)[10],int r)
{
  int i;
  for (i = 0; i < r; i++)
    printf("%s", (*p)[i]);
}

int main()
{
    char *s[10] = { "one", "two", "three", "four" };
    int r;
    r = sizeof(s) / sizeof(s[0]);
    print_2d(&s, r);
}

Note that parameter p is declared as a pointer to an array of char * pointers, which is what your s in main is. However, you have to realize that pointer to array can only point to an array of fixed pre-determined size. The array passed to print_2d will always have size 10 and only 10. For this reason, it makes no sense to pass array size to print_2d separately. Just calculate the size inside and get rid of the second parameter

void print_2d(char *(*p)[10])
{
  int i, r;
  r = sizeof *p / sizeof **p;
  for (i = 0; i < r; i++)
    printf("%s", (*p)[i]);
}

int main()
{
    char *s[10] = { "one", "two", "three", "four" };
    print_2d(&s);
}

Alternatively, your code can be fixed as shown in @BLUEPIXY's answer, but that is a completely different way to interpret your intent. And we don't know what your intent was.

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1 Comment

thanks a lot from last few days i was search lot for the solution how to catch array of pointer with pointer to an array..finally i got thanks a lot :-)
2

Try this

#include<stdio.h>
void print_2d(char *p[10],int r) // This is another change.
{
    int i;
    for(i=0;i<r;i++)
        printf("%s",p[i]); // This is the main change
}

int main() // main has to return int
{
    char *s[10]={"one","two","three","four"};
    int r;
    r=sizeof(s)/sizeof(s[0]);
    print_2d(s,r)
}

s[0], s[1] etc are the strings. Why are you trying to print *s[0], *s[1] etc?

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4 Comments

sorry char*p[10] is an array of pointer actually i want to catch with pointer to an array (*p)[10] and then print them?
@SumitKumar The "main change" is the important thing here. Because of the implicit conversion when passing a parameter, the difference between an array of pointers and a pointer to an array is probably irrelevant. (However, there might be other implications that aren't being addressed by this question.)
@SumitKumar - from main you are passing an array of pointers. Hence you should accept an array of pointers & not pointer to an array.
@user93353 that was only my question i was asked in an interview i have to catch array of pointer with pointer to an array?please can u explain is it possible i said him its not possible
1

The problem you are having is r = 10 every time. So when you pass print_2d(s,r) you are telling print_2d there are 10 strings to print. You cannot get the r using sizeof (arr)/sizeof(arr[0]) it will return 10 each time. You have created a pointer to 10 character arrays with your assignment and initialization (which null terminates the 4 given strings by default and leaving 6 empty). To prevent teh segfault, you need to pass the number of character arrays to print_2d. For example:

#include<stdio.h>

void print_2d(char **p, int r) // i am catching with pointer to an array
{
    int i;
    for (i=0; i<r; i++)
        printf ("%s\n", p[i]);
}

int main (void)
{
    char *s[10] = {"one","two","three","four"};
    int r;
    // r = sizeof(s)/sizeof(s[0]);  // r = 10 causing segfault @ i = 4 in print_2d
    // you cannot get the number of string arrays in this manner. 
    r = 4;
    printf("r: %d\n", r);
    print_2d (s,r);
    return 0;
}

output:

$ ./bin/par
r: 4
one
two
three
four

If you want to use sizeof (s) / sizeof (s[0]) then you will need to declare s as follows:

char *s[] = {"one","two","three","four"};
int r;
r = sizeof(s)/sizeof(s[0]);  // Now r will equal 4
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2 Comments

Though, it's better to not declare size of array in this case - declaring it shouldn't cause an issue - s[4] to s[9] would be null pointers.
No not in this case. Take a 2D array s[row][col]. When your convert to pointer notation *(s[row] + col) as done here, you are creating a pointer to 10 strings with *s[10]. While 4-9 may be empty sizeof(s)/sizeof(s[0]) will still return 10 instead of 4. By declaring *s[] you let the compiler fill in the number of rows from the designation of string literals. (thank you, you helped catch a mistake in my answer)
1

char (*p)[10] is pointer to char[10], not char *s[10] (array of 10's pointer to char)
There is a need to change the main and if void print_2d(char (*p)[10], int r) is correct.

#include <stdio.h>

void print_2d(char (*p)[10], int r){
    int i;
    for(i=0;i<r;i++)
        printf("%s\n", p[i]);
}

int main(){
    char s[][10]={"one","two","three","four"};
    int r;
    r=sizeof(s)/sizeof(s[0]);
    print_2d(s,r);
    return 0;
}
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