在 C++ 中，如何从字符串中获取接下来的几个字符？

Question

My goal is to grab a specific part of a very large number and concatenate that part with another number, then continue. Since integers only go so high, I have a string of the number. I do NOT know what this number could be, so I can't input it in myself. I can use substr for the first part, but I am stuck shortly after.

An example

"435509590420924949"

I want to take the first 5 characters out, convert to integer, do my own calculation to them, then concatenate them with the rest of the string. So I will take 43550 out, do formula to get 49, then add 49 to another 5 in a row after the original string "95904" so the new answer will be "4995904".

This is my code for the first part I made up,

string temp;
int number;

temp = data.substr(0, 5);
number = atoi(temp.c_str());

This grabs the first first characters in the strings, converts to integers where I can calculate it, but I don't know how to grab the next 5 of the long string.

Answer 1

You can get the length of the string, so something like:

std::size_t startIndex = 0;
std::size_t blockLength = 5;
std::size_t length = data.length();
while(startIndex < length)
{
    std::string temp = data.substr(startIndex, blockLength);
    // do something with temp
    startIndex += blockLength;
    // TODO: this will skip the last "block" if it is < blockLength,
    // so you need to modify it a bit for this case.
}

Answer 2

You can use loops. For example:

std::size_t subStrSize = 5;
for (std::size_t k = 0; k < data.size(); k+=subStrSize) {
    std::size_t h = std::min(k + subStrSize - 1, data.size() - 1);
    int number = 0;
    for (std::size_t l = k; l <= h; ++l) 
        number = number * 10 + data[l] - '0';
    //-- Some work with number --
}