c++ pass array directy to function - c++

Question

bool foo(const char* arg, const char* str[]){
for (int i=0; i < (signed)sizeof(str); i++){
if(strcmp(arg, str[i])) == 0){
return true;
}
}
return false;
}
int main(){
foo("c", {"a", "b", "c"});
return 0;
}
I want to pass an array of const char* to the function directly like it is shown here. But I keep getting the error: cannot convert '<brace-enclosed initializer list>' to 'const char**'

Answer 1

While there is a way to make that work using templates and passing C-arrays by reference, I'm not sure if that is what you really want: example
In C++11 I'd much prefer to use a std::intializer_list to make the call foo("c", {"a", "b", "c"}) work as intended:
#include <initializer_list>
bool foo(const char* arg, std::initializer_list<const char*> strings) {
for (const char* str : strings) {
/* ... */
}
}
While we're at it, you should consider using C++'s string facilities over those inherited from C:
#include <initializer_list>
#include <string>
bool foo(const std::string& arg, std::initializer_list<std::string> strings) {
for (const auto& str : strings) {
if (arg == str) {
return true;
}
}
return false;
}
Note that a std::initializer_list does not own the values it provides access to, so if you want to store it, use a proper container like std::vector instead.
Also, if checking for the presence of arg in str is all you want to do,
why not use std::find?
#include <algorithm>
#include <initializer_list>
#include <string>
bool foo(const std::string& arg, std::initializer_list<std::string> strings) {
return std::find(strings.begin(), strings.end(), arg) != strings.end();
}

Answer 2

Use a character pointer pointer (char**) to achieve this. You cannot declare parameters as arrays, but you can pass arrays into functions, as they will be converted into double pointers.

Answer 3

You can also pass the array by reference via a template function, so you will be able to detect its size at compile time (note that in your case the array decays to a pointer, so sizeof doesn't do what you think). Example:
#include <iostream>
#include <cstring>
template<int N>
bool foo(const char* arg, const char* (&str)[N])
{
for (std::size_t i = 0; i < N; i++) {
if (std::strcmp(arg, str[i]) == 0) {
return true;
}
}
return false;
}
int main() {
const char* arr[] = {"abc", "de"};
std::cout << std::boolalpha << foo("de", arr);
}
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And btw, try to use algorithms from the C++ standard library and std::string instead of strcmp() and raw C-strings:
#include <algorithm>
#include <iostream>
#include <vector>
bool foo(const std::string arg, const std::vector<std::string>& str)
{
return std::find(str.begin(), str.end(), arg) != str.end();
}
int main()
{
std::cout << std::boolalpha << foo("de", {"abc", "de"});
}
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Answer 4

Although it is recommended to drop C-style programming when using C++, and use stl containers instead, I would like to mention that the following "one-liner" works:
bool foo(const char* arg, const char* const* str, size_t siz) {
for (size_t i = 0; i < siz; i++)
if (!strcmp(arg, str[i])) return true;
return false;
}
int main() {
cout << foo("c", array<char*, 3> { "a", "b", "c" }.data(), 3);
system("pause");
}
The surprising surprise is that if we define the second argument of foo as const char**, the compiler (VS2015) yells and says:
bool foo(...): cannot convert argument 2 from 'char **' to 'const char **'
:O
cannot convert from non-const to const??? Does that make sense anyone?
Edit: #NeilKirk explained it in his comment. Normal. Accordingly the signature should be as:
bool foo(const char* arg, const char* const* str, size_t siz)
and everything works as expected.