I need to convert a part of vector
s of char
s to an int
.
If I have
std::vector<char> // which contains something like asdf1234dsdsd
and I want to take characters from the 4th till 7th position and convert it into an int.
The positions are always known.
How do I do it the fastest way ?
I tried to use global copy and got a weird answer. Instead of 2 I got 48.
如果位置已知,您可以这样操作
int x = (c[4] - '0') * 1000 + (c[5] - '0') * 100 + (c[6] - '0') * 10 + c[7] - '0';
复制不起作用的原因可能是由于字节顺序,假设第一个char
是最重要的:
int x = (int(c[4]) << 24) + (int(c[5]) << 16) + (int(c[6]) << 8) + c[7]
This is fairly flexible, although it doesn't check for overflow or anything fancy like that:
#include <algorithm>
int base10_digits(int a, char b) {
return 10 * a + (b - '0');
}
int result = std::accumulate(myvec.begin()+4, myvec.begin()+8, 0, base10_digits);
1.Take the address of the begin and end (one past the end) index.
2.Construct a std::string from it.
3.Feed it into a std::istringstream.
4.Extract the integer from the stringstream into a variable.
(This may be a bad idea!)
Try this:
#include <iostream>
#include <vector>
#include <math.h>
using namespace std;
int vtoi(vector<char> vec, int beg, int end) // vector to int
{
int ret = 0;
int mult = pow(10 , (end-beg));
for(int i = beg; i <= end; i++) {
ret += (vec[i] - '0') * mult;
mult /= 10;
}
return ret;
}
#define pb push_back
int main() {
vector<char> vec;
vec.pb('1');
vec.pb('0');
vec.pb('3');
vec.pb('4');
cout << vtoi(vec, 0, 3) << "\n";
return 0;
}
long res = 0;
for(int i=0; i<vec.size(); i++)
{
res += vec[i] * pow (2, 8*(vec.size() - i - 1)); // 8 means number of bits in byte
}
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