I've seen some videos where a 2D array is created to store the strings, but I wanted to know if it is possible to make a 1D array of strings.
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1stackoverflow.com/a/27705098/17856705Computable– Computable2022-07-11 23:47:20 +00:00Commented Jul 11, 2022 at 23:47
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1 Answer
NOTE: In C, a string is an array of characters.
//string
char *s = "string";
//array of strings
char *s_array[] = {
"array",
"of",
"strings"
};
Example
#include <stdio.h>
int main(void)
{
int i = 0;
char *s_array[] = {
"array",
"of",
"strings"
};
const int ARR_LEN = sizeof(s_array) / sizeof(s_array[0]);
while (i < ARR_LEN)
{
printf("%s ", s_array[i]);
i++;
}
printf("\n");
return (0);
}
answered Jul 11, 2022 at 23:50
Maxwell D. Dorliea
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5 Comments
Gabriel Staples
It might be useful to make a full, runnable example as well and show how to print all strings in the array.
chux
"In C, a string is an array of characters." --> better as "In C, a string is an array of characters with a terminating null character.". C lib defines it as: "A string is a contiguous sequence of characters terminated by and including the first null character."
chux
Instead of
const int ARR_LEN = 3;, could determine the count from s_array{} with const int ARR_LEN = sizeof s_array / sizeof s_array[0];.Gabriel Staples
Like @chux-ReinstateMonica said, you can get the array length from the array. Here is a macro I like to use:
#define ARRAY_LEN(array) (sizeof(array) / sizeof(array[0])). Example usage: search this file for ARRAY_LEN(: array_2d_practice.c.
ad absurdum
s isn't a string, it is a pointer to (the first element of) a string. Similarly, s_array isn't an array of strings, it is an array of pointers. One reason that this matters is because attempts to modify the strings referenced by s or by the pointers in s_array lead to undefined behavior. You could create a string with char s[] = "string";, or an array of strings with char s_array[][4] = { "abc", "123" };. These strings can be modified.