Substring exists in string in C - c

Question

#include <stdio.h>
#include <string.h>
int main()
{
char str1[80] = "downtown", str2[20] = "town";
int len1 = 0, len2 = 0, i, j, count;
len1 = strlen(str1);
len2 = strlen(str2);
for (i = 0; i <= len1 - len2; i++) {
for (j = i; j < i + len2; j++) {
count = 1;
if (str1[j] != str2[j - i]) {
count = 0;
break;
}
}
if (count == 1) {
break;
}
}
if (count == 1) {
printf("True");
} else {
printf("False");
}
}
In the above code, I'm trying to solve this one without using string functions apart from strlen() which can be replaced with a simple while loop. Is there any other way of checking for consecutive characters like firstly checking if the character is in the string, and if the i index is in the next position and not randomly in the string.

Answer 1

Here is a very clean way to do it using a function. It assumes that both str and sub are proper C-strings and returns a pointer to first match and NULL if no match.
char *substr(const char *str, const char *sub) {
if (!*sub)
return str; // Empty string is substring of all strings
while (*str) {
const char *sub1 = sub;
const char *str1 = str;
while (*str1++ == *sub1++) {
if (!*sub1)
return (char *)str;
}
str++;
}
return NULL;
}
This function is identical to the standard function strstr(), present in the C Standard library and declared in <string.h>.

Answer 2

There are some problems in the posted code:
count is uninitialized and only set if the inner loop is reached, which will not be the case if len1 < len2. count should be intialized to 0 to handle this case properly. found would be a more informative name for this variable.
Furthermore, the statement count = 1; should be moved before the inner loop to handle the case of an empty substring.
Here is a modified version:
#include <stdio.h>
#include <string.h>
int main() {
char str1[] = "downtown", str2[] = "town";
int found = 0;
size_t len1 = strlen(str1);
size_t len2 = strlen(str2);
if (len1 >= len2) {
for (size_t i = 0; i <= len1 - len2; i++) {
found = 1;
for (size_t j = 0; j < len2; j++) {
if (str1[i + j] != str2[j]) {
found = 0;
break;
}
}
if (found) {
break;
}
}
}
if (found) {
printf("True\n");
} else {
printf("False\n");
}
return 0;
}

Answer 3

I did not readily see OP's error. See #chqrlie.
Minor: Use size_t i, not int to cope with long strings.
strlen(str1) runs down the entire string, even when not needed, so that is avoided.
Alternative:
Fixed a faulty case in #klutt otherwise good answer.
Added const for greater application.
Return beginning of match on success.
Simplified and test harness added.
I like the needle in a haystack identifiers
char* substr3(const char *haystack, const char *needle) {
while (*haystack) {
const char *htmp = haystack;
const char *ntmp = needle;
while (*htmp == *ntmp && *htmp) {
htmp++;
ntmp++;
}
if (!*ntmp) {
return (char*) haystack; // Beginning of match
}
haystack++;
}
return *needle ? NULL : (char *) haystack;
}
int main(void) {
printf("%s\n", substr3("ababc", "abc"));
printf("%s\n", substr3("abc", "abc"));
printf("%s\n", substr3("abd", "ab"));
printf("%s\n", substr3("abc", ""));
printf("%s\n", substr3("", ""));
printf("%p\n", substr3("a", "abc"));
printf("%p\n", substr3("aba", "abc"));
printf("%p\n", substr3("x", "ab"));
printf("%p\n", substr3("aaa", "ab"));
return 0;
}
Some tighter code
char* substr4(const char *haystack, const char *needle) {
do {
const char *htmp = haystack;
const char *ntmp = needle;
while (*htmp == *ntmp && *ntmp) {
htmp++;
ntmp++;
}
if (!*ntmp) {
return (char*) haystack; // Beginning of match
}
} while (*haystack++);
return NULL;
}

Answer 4

Simple && naive:
char *substr2(char *str, char *sub) {
if (!*sub) return str; // not needed: see the generated code.
for(; *str; str++) {
size_t pos;
for(pos=0; str[pos] ; pos++) {
if (str[pos] != sub[pos]) break;
}
if (!sub[pos]) return str;
}
return NULL;
}
Generally:
if you don't try to outsmart the compiler: you win.
fewer variables: you win
fewer conditions inside the loop: win
try to be smart: you'll lose
when all else fails: use KMP or BM search