如何将字符串数组指针传递给新的字符数组？ 在 C

Question

I have an array pointer and I try to pass strings of this pointer to a new array. However, I can not print out the new array. My code is:

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

int main() {
    char *array[5] = {"may", "june", "july", "august", "september"};
    char buffer[5];
    for (int i=0; i<5; i++) {
        buffer[i] = *array[i];
        printf(%s , buffer[i]);     
    }
}

I can not compile the program because it got error of %s. How can I print out the buffer array like:

may june july august september

without changing it format char buffer[5] which means I do not want to change it to *buffer[5] or anything else. Thank you.

Answer 1

Not sure exactly what you're wanting to do. You can use strcpy() to copy the contents of one array to another.

char* array[5] = {"may", "june", "july", "august", "september"};
char arr[10];

for(i = 0; i < 5; ++i) {
    strcpy(arr, array[i]);
    printf("%s\n", arr);
}

Or if you want to do it like this instead with a 2d char array.

char* array[5] = {"may", "june", "july", "august", "september"};
char arr2d[5][10];

for(i = 0; i < 5; ++i) {
    strcpy(arr2d[i], array[i]);
    printf("%s\n", arr2d[i]);
}

Answer 2

I can not compile the program because it got error of %s. How can I print out the buffer array like:

 may june july august september

Quite simply print out each index of array , eg

#include <stdio.h>

int main() {

    char *array[] = {"may", "june", "july", "august", "september"};
    int nmembers = sizeof array / sizeof *array;

    for (int i = 0; i < nmembers; i++)
        printf(" %s", array[i]); 
    putchar ('\n');     /* tidy up with newline */
}

note: do not use magic numbers in your code. There is no need. You can compute the number of elements in array using sizeof (only when array is declared within the current scope)

Example Use/Output

$ ./bin/parraymess
 may june july august september

Why does it work that way?

char *array[] = {"...", ...}; declares an array of pointers to char and each pointer making up the array is initialized to point to the beginning of each string-literal used within the initializer., eg

array[0] = "may";
array[1] = "june";
...

Since you have an array of pointers to char , each element is a pointer to char , eg char* . When you access each individual member of an array using name[..] the [..] acts as a dereference . So simply looping over each element for (int i = 0; i < nmembers; i++) and then accessing array[i] references a pointer (eg char* ) to the beginning of each string that can be used with the "%s" format specifier in printf .

Look things over and let me know if you have further questions.

Answer 3

You cannot store a string in space of a char . You can either have an array of char * for a string (and allocate memory dynalically) or you can store the strings in a two-dimensional char array eg:

char buffer[5][MAX_LEN];

and then you may copy strings from array to buffer using either strcpy or by writing your own function.

Answer 4

How about this:

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

int main(void) {
    char *array[5] = {"may", "june", "july", "august", "september"};
    char *buffer[5];
    int i;
    for (i=0; i<5; i++) {
        buffer[i] = strdup(array[i]);
        printf("%s ", buffer[i]);
    }
    return 0;
}

You need buffer to be an array of 5 pointers to char. You have declared buffer as a char array of size 5 , this will not be large enough to hold strings larger than 4 chars (mind the trailing \\0 ).

strdup is a function which takes a c-string, and writes that string to some space it allocates by calling malloc and returns a pointer to that new space.

If you really need buffer be a char array, you should make sure it can hold the longest string you want to write to it and a trailing \\0 . Then you could use the strcpy function to write data to buffer . Like so:

int main(void) {
    char *array[5] = {"may", "june", "july", "august", "september"};
    char buffer[12];
    int i;
    for (i=0; i<5; i++) {
        strcpy(buffer, array[i]);
        printf("%s ", buffer);
    }
    return 0;
}