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I have the following code:
void test_func(const char(*str)[4])
{
std::cout << *str << std::endl;
}
caller code:
std::string s{"abc"};
char c[4];
strcpy_s(c, s.c_str());
test_func(&c);
It is possible to replace char[4] with modern std::vector<char> ? If it is, how can I do that ?
I have tried a silly code:
std::vector<char> chars(s.cbegin(), s.cend());
test_func(&chars[0]);
Of course, didn't work it.
A high-level, general-purpose programming language, created as an extension of the C programming language, that has object-oriented, generic, and functional features in addition to facilities for low-level memory manipulation.
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Depending on what you want, have a look at using a string_view or span of char as the parameter to your function.
I don't understand what you are trying to accomplish here. You could code test_func to take a const char* pointer.
The code suppose to be compiled as c++11. So, I don't think I can use your choices ... am I ?
I have tried:
test_func("123");
but my compiler respond me:
error C2664: 'void test_func(const char (*)[4])': cannot convert argument 1 from 'const char [4]' to 'const char (*)[4]'
No, you can't. As you'd mentioned "modern", I presumed it was modern C++ :)
I still don't understand the point of the exercise. By design, a C-style array can be passed to a function taking a const char* pointer.
I am using a cpp library which has a function with that proptotype:
void test_func(const char(*str)[4])
Yes, I could use strcpy, but I want to modernize the code.
Answer accepted by question author
It's not safe but you could call the library function using a properly sized string s -
test_func(reinterpret_cast<const char(*)[4]>(s.c_str()));