Question

我的test.cpp包含以下代码。我用valgrind描述了它，valgrind显示错误。我错过了什么？

#include<string>
#include<cstring>
#include<iostream>

void TestString(std::string & str)
{
    char * tmpBuff = new char[3];
    tmpBuff[0] = 'f';
    tmpBuff[1] = 'o';
    tmpBuff[2] = 'o';
    str = tmpBuff;
    delete [] tmpBuff;
}

int main(int argc, char* argv[])
{
    std::string test_string;
    TestString(test_string);
    std::cout<<test_string;
    return 0;
}

的valgrind日志

==5026== Invalid read of size 1
==5026==    at 0x4A07F64: strlen (mc_replace_strmem.c:403)
==5026==    by 0x347E29E14B: std::string::operator=(char const*) (in /usr/lib64/libstdc++.so.6.0.13)
==5026==    by 0x4009AD: TestString(std::string&) (test.cpp:11)
==5026==    by 0x4009EC: main (test.cpp:18)
==5026==  Address 0x4c22043 is 0 bytes after a block of size 3 alloc'd
==5026==    at 0x4A07152: operator new[](unsigned long) (vg_replace_malloc.c:363)
==5026==    by 0x400979: TestString(std::string&) (test.cpp:7)
==5026==    by 0x4009EC: main (test.cpp:18)

Answer 1

tmpBuff缺少终止\0。

它应包含4个字符：'f', 'o', 'o', '\0'。

Answer 2

尝试以下修复

void TestString(std::string & str)
{
    char * tmpBuff = new char[4]; // <<<
    tmpBuff[0] = 'f';
    tmpBuff[1] = 'o';
    tmpBuff[2] = 'o';
    tmpBuff[3] = '\0'; // <<<
    str = tmpBuff;
    delete [] tmpBuff;
}

C风格的字符串需要终止\0字符。

Answer 3

您呼叫的std::string(const char*)构造函数需要nul terminated string。您没有传递一个，因此结果是未定义的行为。构造函数将尝试读取，直到找到\0。

所以，传递一个nul终止的字符串，一切都会好的。

读取大小1无效，将char *转换为std :: string

3 个答案: