Find string inside 2D char array in C - c++

Question

I am trying to find a string which is inside 2D char array and return it's index. For example:
char idTable[255][32];
char tester[] = { 't','e','s','t','e','r','\0' };
memcpy(idTable[43], tester, 7);
uint8_t id = getID(name[0]);
//name is returned from function "char **name = func();"
//but I have the same results when I try using normal char array...
I've had partial success with the first part of the below code, but it is finding a match if a part of the word is the same (one, oneTwo). If I add "else if" to the first "if" it always goes to the "else if".
The rest of the file prints different results for
printf("idTable string lenght:\t %u\n", strlen(idTable[index]));
and
printf("foundMatch string lenght:\t %u\n", strlen(foundMatch));
, unless I add printf("Index:\t %i\n", index);.
uint8_t getID(char *name) {
printf("\nInserted name:\t %s\n", name);
uint8_t index;
for (uint8_t r = 0; r < 255; r++) {
if (strstr(idTable[r], name) != NULL) {
printf("Found '%s' in position:\t %d\n", name, r);
index = r;
}
}
printf("Index:\t %i\n", index); // THIS LINE
char foundMatch[strlen(idTable[index])];
printf("idTable string lenght:\t %u\n", strlen(idTable[index]));
for (uint8_t c=0; c<strlen(idTable[index]); c++) {
foundMatch[c] = idTable[index][c];
}
printf("foundMatch string lenght:\t %u\n", strlen(foundMatch));
if (strcmp(foundMatch, nodeName) == 0) {
printf("Confirmed\n");
return index;
} else {
printf("Second test failed\n");
return 0;
}
}
Why am I getting this strange results and is there a better way to do this?

Answer 1

I don't know how you are initializing your idTable entries, but if you are using the method that you showed at the start of the question you'll have problems. You can't assume all of the space reserved by idTable is initialed to 0's, so idTable[43] isn't a null terminated string. Therefore idTable[43] need not compare equal to the null terminated string "tester".
Also your getID function doesn't return anything despite its signature. So it won't even compile as-is.

Answer 2

Here's a solution in actual C++, not C.
std::array<std::string, 255> idTable;
idTable.at(43) = "tester";
std::pair<std::size_t, std::size_t> findInIdTable(std::string const& what) {
for (unsigned i = 0; i < idTable.size(); ++i) {
std::size_t pos = idTable.at(i).find(what);
if (pos != std::string::npos) {
return std::make_pair(i, pos);
}
}
// if the code reaches this place, it means "not found". Choose how you want to deal with it
// my personal suggestion would be to return std::optional<std::pair<...> instead.
}
If you want to discard the pos value, it's easy to change as well.
Live On Coliru

Answer 3

In the category: Use C++
Of course, use std::array<char, 32> or std::string if possible. I stuck with your choices for this answer:
Live On Coliru
#include <algorithm>
#include <iostream>
#include <cstring>
char idTable[255][32] = { };
int main() {
using namespace std;
// initialize an entry
copy_n("tester", 7, idTable[43]);
// find match
auto match = [](const char* a) { return strcmp(a, "tester") == 0; };
auto index = find_if(begin(idTable), end(idTable), match) - idTable;
// print result
cout << "match at: " << index;
}
Prints
match at: 43

Answer 4

You need to add a nul to the end of the foundMatch array after copying in the idTable row:
foundMatch[strlen(idTable[index])] = '\0';
right before the 'foundMatch string lenght' (length) message.
strlen is an expensive function that walks the string every time. You should call that once, store it in a local variable, then reference that variable rather than calling strlen repeatedly.