使用strcpy()与动态内存

Question

我的代码运行正常，没有内存泄漏。但是，我收到了valgrind错误：

==6304== 14 errors in context 4 of 4:
==6304== Invalid write of size 1
==6304==    at 0x4A0808F: __GI_strcpy (mc_replace_strmem.c:443)
==6304==    by 0x401453: main (calc.cpp:200)
==6304==  Address 0x4c390f1 is 0 bytes after a block of size 1 alloc'd
==6304==    at 0x4A075BC: operator new(unsigned long) (vg_replace_malloc.c:298)
==6304==    by 0x401431: main (calc.cpp:199)

==6304== 4 errors in context 2 of 4:
==6304== Invalid read of size 1
==6304==    at 0x39ADE3B0C0: ____strtod_l_internal (in /lib64/libc-2.12.so)
==6304==    by 0x401471: main (calc.cpp:203)
==6304==  Address 0x4c390f1 is 0 bytes after a block of size 1 alloc'd
==6304==    at 0x4A075BC: operator new(unsigned long) (vg_replace_malloc.c:298)
==6304==    by 0x401431: main (calc.cpp:199)

错误1和3分别与2和4相同，除了初始地址以外。

这些错误是什么意思，我该如何修复它们？

 int main(){

    //Dlist is a double ended list. Each Node has a datum,
    //a pointer to the previous Node and a pointer to the next Node
    Dlist<double> hold;
    Dlist<double>* stack = &hold;

    string* s = new string;
    bool run = true;
    while (run && cin >> *s){

        char* c = new char;
        strcpy(c, s->c_str());   //valgrind errors here

        if (isdigit(c[0]))
            stack->insertFront(atof(c));

        else{
            switch(*c){
                //calculator functions
            }

        delete c;
        c = 0;
  }
delete s;
s = 0;

- user2961535

---

那问题是什么？ - juanchopanza

你为什么要动态分配内存呢？在上述应用程序中没有必要这样做。 - Martin York

当您修复代码后，最好在codereview.stackexchange.com上发布帖子，以获取有关实现目标更好方式的意见。 - Martin York

Answer 1

Valgrind会发出许多通常无害的警告，因为stdlib函数有一些小技巧，但是这里不是这种情况:

char* c = new char; // this is bad

只分配一个 char，而不是 char 缓冲区，请尝试：

char* c = new char[s->size()+1];

然后将删除更改为：

delete [] c;

- Glenn Teitelbaum

Answer 2

char* c = new char; c的大小为1，即使要复制一个字符的字符串，也需要两个字符长的缓冲区（第二个字符用于保存空终止符）

- asifgoldpk

Answer 3

就在这里：

    char* c = new char;

你只分配了一个字符。应该分配一个数组：

    char* c = new char[str->length() + 1];

请记得调用delete[]。您分配了+1的空间来容纳字符串的空结尾。

- Shaggi

Answer 4

    char* c = new char;

您正在分配一个单个字符，然后将一个过长的字符串复制到该内存中，无法容纳。您需要分配足够大的数组。

- Graham Borland