因此,我为我在课堂上编写的ADT编写了这两个版本的重载赋值运算符。当我将它们与同一个ADT的重载的ostream <
void Text::operator= (const Text &other) {
if (this != &other) {
delete [] buffer;
bufferSize = other.bufferSize;
buffer = new char[bufferSize + 1];
strcpy(buffer, other.buffer);
}
}
void Text::operator= (const Text &other) {
if (this != &other) {
delete [] buffer;
bufferSize = other.bufferSize;
buffer = new char[bufferSize + 1];
for (int i = 0; i < bufferSize; i++) {
buffer[i] = other.buffer[i];
}
}
这是我超载的ostream <
ostream & operator << (ostream &output, const Text &outputText) {
output << outputText.buffer;
return output;
}
出现这种差异是这样的:
第一个输出:
Hey Jude
第二个输出:
Hey Jude(random garbage)
最佳答案
第二个代码段未附加终止的空终止符,因此出现了垃圾(strcpy()
复制了终止的空终止符)。您需要在for
循环后显式添加空终止符:
buffer[bufferSize] = 0;