Thank you. Today I have a string called "abdcsk". What should I do if I want to type another string to see whether the character on the front side only shows the "abdcsk" face (that is, there will be no fwe and so on)? can I actually do it in strstr? thank you!
if the granularity of your comparison is a single character, not a whole string, then strstr will not work. It's better to use a double loop, a char and a char to judge for yourself.
use the Set bar of es6
to find the union set and judge the size
const s1 = 'abcsk'
const s2 = 'abnn'
if (new Set(s1).size === new Set(s1 + s2).size) {
// s2s1
} else {
// s2s1
} can be handled using the following library functions:
size_t strspn (const char str1, const char str2);
-sharpinclude <stdio.h>
-sharpinclude <stdlib.h>
-sharpinclude <string.h>
int main()
{
const char* str1 = "ab2dcskabdcskabdcskabdcskabdcsk";
const char* str2 = "abdcsk";
size_t l = strspn(str1, str2);
if (l == strlen(str1)) {
printf("All characters in str1:<%s> are all in str2:<%s>\n", str1, str2);
} else {
printf("The first %d characters in str1:<%s> are in str2:<%s>\n", l, str1, str2);
}
return 0;
}Previous: Array object reordering
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