从函数填充的字符串的推导向量中的字符串到int

Question

I have a string 111;222;333 which I want to convert to three integers.

First I split the string

std::vector<std::string> split(...){ ... };

The return values are stored in vector of deduced type

std::vector splitVals {split(...)};

If I then want to convert the values to integer like so

int foo1 {std::stoi(splitVals[0])};

The stoi function is complaining, because the deduced type of the vector is std::vector<std::vector<std::string>, std::allocator<std::vector<std::string>>> , but if I don't let the type to be deduced, everything works as intended.

std::vector<std::string> splitVals {split(...)};
int foo1 {std::stoi(splitVals[0])};

std::stoi now can work because the input value is std::string . The issue seems to start with the vector being initialized from a function which returns the std::vector<std::string> . Is there a way to benefit from C++17 class template argument deduction, without these limitations?

Answer 1

Don't use brace initialization. It favors the std::initializer_list constructor, and so the vector is deduced as holding sub-vectors for elements. Using parentheses should clear it right up.

std::vector splitVals (split(...));

Now the copy deduction guide is favored and the type held by the vector should be deduced as std::string .

Answer 2

Wow. 8 upvotes and accepted answer. So, basically nothing more to add . . .

I just want to come back to the underlying mentioned task:

I have a string 111;222;333 which I want to convert to three integers.

This a typical example for parsing for a CSV. And this can be easily done with a one-liner (one statement). Using std::transform and sregex_toke_iterator .

For completeness, I will show the code for that.

#include <iostream>
#include <vector>
#include <string>
#include <regex>
#include <iterator>
#include <algorithm>

int main()
{
    // The test data
    std::string data("111;222;333;444");

    // Here we will see the result
    std::vector<int> values{};

    // This we will look up in the string
    std::regex re("(\\d+)");

    // Put all tokens into vector
    std::transform(
        std::sregex_token_iterator(data.begin(), data.end(), re, 1), 
        std::sregex_token_iterator(),
        std::back_inserter(values),
        [](const std::string& s){ return std::stoi(s); }
    );

    // Show debug output    
    std::copy(values.begin(),values.end(),std::ostream_iterator<int>(std::cout,"\n"));

    return 0;
}