我重載operator*
爲std::string
類,但在這種情況下:爲什麼在運算符重載時會發生自由錯誤?
std::string operator*(std::string a, unsigned b) //bad
{
unsigned old_length = a.length();
a.resize(a.length()*b);
for(unsigned i = old_length ;i<a.length()*b; i++)
a[i]=a[i%old_length];
return a;
}
程序錯誤崩潰:
*** Error in `./program': free(): invalid next size (fast): 0x0000000000cd20b0 *** Aborted
如果我重載它像這樣 - 沒有錯誤:
std::string operator*(std::string a, unsigned b)
{
unsigned old_length = a.length();
std::string a2 = a;
a2.resize(a.length()*b);
for(unsigned i = 0 ;i<a.length()*b; i++)
a2[i]=a[i%old_length];
return a2;
}
那麼問題在哪裏?有沒有辦法創建新的字符串a2
?它消耗更多的內存。
#include <iostream>
#include <string>
std::string operator*(unsigned b, std::string a)
{
return operator*(a, b);
}
int main(int argc, char **argv)
{
std::string a = "abcdef "; // if string contains more than 4 symbols - free error for the first case
std::string aaaa = 4*a;
std::cout << a << "\n"
<< aaaa << "\n"
<< std::endl;
return 0;
}
'自動N =則爲a.length(); a.reserve(n * b); while(b--)a.append(a,0,n); return a;'應該更好... – Deduplicator
確實更好:) – LogicStuff
有一些'const'在那裏,是啊 –