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I am noob and I want to pass pointer of an array of strings to a function and that function simply prints the strings of the array using pointer arithmetic.

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

char words[][10] = { "madam", "Russia", "India", "Japan", "China", "Brazil", "Mexico" };
int size = sizeof(words) / sizeof(words[0]);

void printer(char *, int);

int main() {
    printer(*words, size);

    return 0;
}

void printer(char *array, int len) {
    for (int i = 0; i < len; i++) {
        printf("%s\n", (array + i));
    }
}

Output:

madam
adam 
dam  
am   
m

It is not printing elements (strings) one by one.

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2
  • 1
    Funny question. When you tried that, you must have written a function main() (well, you didn't) that does exactly what you want - Why did you try something different in your printer function? Commented Apr 25, 2022 at 7:42
  • The words, do you need it to be array of arrays of 10 characters which is what you have now (can fit 9 character string)? Or would you be find with array of pointer to const char (which can be string literal), that is const char *words[] = { ... };? Commented Apr 25, 2022 at 7:59

3 Answers 3

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2

As coded, printer takes a string and prints every final substring. You pass *words, which is the same as words[0], ie: the first string in the 2D array.

To pass a pointer to the array, the type of the argument should be char array[][10](*).:

#include <stdio.h>

char words[][10] = { "madam", "Russia", "India", "Japan", 
                     "China", "Brazil", "Mexico" };
int size = sizeof(words) / sizeof(words[0]);

void printer(char words[][10], int len);

int main() {
    printer(words, size);
    return 0;
}

void printer(char words[][10], int len) {
    for (int i = 0; i < len; i++) {
        printf("%s\n", array[i]));
    }
}

Note however that you could also define array as an array of pointers to character strings char *array[]. This would allow for strings of any length, not limited to 9 characters plus a null terminator.

The code would be changed to this:

#include <stdio.h>

char *words[] = { "madam", "Russia", "India", "Japan", 
                  "China", "Brazil", "Mexico" };
int size = sizeof(words) / sizeof(words[0]);

void printer(char *words[], int len);

int main() {
    printer(words, size);
    return 0;
}

void printer(char *words[], int len) {
    for (int i = 0; i < len; i++) {
        printf("%s\n", array[i]));
    }
}

The array words is initialized from string literals, which must not be changed, so it would be better to define it as const char *array[] and modify printer accordingly. Note also that printf("%s\n", array[i])) is equivalent to puts(array[i]) (good compilers detect this equivalence and generate the call to puts in both cases):

#include <stdio.h>

const char *words[] = { "madam", "Russia", "India", "Japan", 
                        "China", "Brazil", "Mexico" };
int size = sizeof(words) / sizeof(words[0]);

void printer(const char *words[], int len);

int main() {
    printer(words, size);
    return 0;
}

void printer(const char *words[], int len) {
    for (int i = 0; i < len; i++) {
        puts(array[i]);
    }
}

(*) The function does not receive an actual 2D array as argument, but a pointer to the array of arrays, which technically has type char (*)[10] but the compiler will accept both syntaxes equivalently and char array[][10] is simpler and more readable

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1 Comment

This is a bit dangerous, because you have array of pointers to non-const char, but they point to string literals, which normally are in read-only memory segment. It'd be better to have const char *words[] = { ... };
1

You are just printing the 1st element of words[0], because you have passed words[0] which is equivalent to *words.

Also, to print the whole array of pointers, you need char [][10] as your parameter in function printer().

Note: You can put printer() function above main() function, then there will be no need for forward declaration.

Final Code

#include <stdio.h>

char words[][10] = {"madam", "Russia", "India", "Japan", "China", "Brazil", "Mexico"};
int size = sizeof(words) / sizeof(*words);

void printer(const char array[][10], int len){
    for(int i = 0; i < len; i++){
        puts(array[i]);
    }
}

int main(void) {
    printer(words, size);
    return 0;
}
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3 Comments

Your variable words is an array of pointer. Not at all: array is an array of arrays of 10 bytes.
size_t is not necessarily unsigned long. As a matter of fact it is unsigned int on most 32-bit systems and unsigned long long on 64-bit windows. Avoid confusing newbies with such fallacies. While size_t is indeed the correct type for indices, using int is no problem for most arrays, only huge ones require size_t.
indeed unsigned long on 64-bit linux, but not every machine is a linux PC :)
-2

char *array to char (*array)[10]

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2 Comments

Please explain why the first idiom is not correct.
A code-only answer is not high quality. While this code may be useful, you can improve it by saying why it works, how it works, when it should be used, and what its limitations are. Please edit your answer to include explanation and link to relevant documentation.