Overall arrays in C are very confusing to me, so I have absolutely no idea of how to do this. Here is an example of what I am trying to do.
string hello = "hello";
string array[20] = {"e", "cat", "tree", "hello"};
for (int i = 0; i < 3; i++) {
if (!strcmp(array[i], hello)) {
printf("Hello is in the array");
}
}
Your issue is in the for loops exit condition, your loops stops before i == 3, which is the last index of your array. Since it never hits that, it never compares your string against the last element, where the "hello" is.
#include <stdio.h>
#include <string.h>
typedef char * string;
int main(int argc, char *argv[]) {
string hello = "hello";
string array[20] = {"e", "cat", "tree", "hello"};
for (int i = 0; i < 4 /* not 3 */; i++) {
if (!strcmp(array[i], hello)) {
printf("Hello is in the array");
}
}
}
And you just learned, first hand, why you should never hard code the count of an array like this. It's inevitable that you'll edit your array, and forget to update the hard-coded count in all the places where you used it.
Try this, instead:
#include <stdio.h>
#include <string.h>
typedef char * string;
int main(int argc, char *argv[]) {
string hello = "hello";
string array[] = {"e", "cat", "tree", "hello"};
size_t array_count = sizeof(array) / sizeof(*array);
for (int i = 0; i < array_count; i++) {
if (!strcmp(array[i], hello)) {
printf("Hello is in the array");
}
}
}
array_count is calculated by dividing the size of array, which is an array of four pointers to char, with the size of *array, which is one pointer to char.
string? Is it justtypedef char * string;?cs50.hstringis indeed achar *, so probably that.< 3to<= 3or< 4. I assume this is just a typo, right?