c - LeetCode : Address Sanitizer Violations

Question

我正在研究一些示例，但不断收到这些 Address Sanitizer 堆缓冲区溢出错误。我一生都无法弄清楚潜在的溢出在哪里:

bool detectCapitalUse(char * word){
    
    int CapitalLet = 0;
    int WordLen = sizeof(word)/sizeof(char);
    bool result = 0;
    
    for (int i = 0; i < WordLen; i++)
    {
        if (word[i] >= 'A' && word[i] <= 'Z' )
        {
            CapitalLet++;
        }
    }
    
    if( CapitalLet == WordLen )
    {
        result = 1;
    }
    else if ((CapitalLet == 1) && (word[0] >= 'A' && word[0] <= 'Z'))
    {
        result = 1;
    }
    else if (CapitalLet == 0 )
    {
        result = 1;
    }
    else
    {
        result = 0;
    }    
    
    return result;
}

编辑:这是完整的错误消息。

=================================================================
==31==ERROR: AddressSanitizer: heap-buffer-overflow on address 0x602000000017 at pc 0x000000401850 bp 0x7ffcd5e90680 sp 0x7ffcd5e90670
READ of size 1 at 0x602000000017 thread T0
    #2 0x7f2bb38ea82f in __libc_start_main (/lib/x86_64-linux-gnu/libc.so.6+0x2082f)
0x602000000017 is located 0 bytes to the right of 7-byte region [0x602000000010,0x602000000017)
allocated by thread T0 here:
    #0 0x7f2bb4905f88 in malloc (/usr/lib/x86_64-linux-gnu/libasan.so.5+0x10bf88)
    #4 0x7f2bb38ea82f in __libc_start_main (/lib/x86_64-linux-gnu/libc.so.6+0x2082f)
Shadow bytes around the buggy address:
  0x0c047fff7fb0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
  0x0c047fff7fc0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
  0x0c047fff7fd0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
  0x0c047fff7fe0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
  0x0c047fff7ff0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
=>0x0c047fff8000: fa fa[07]fa fa fa fa fa fa fa fa fa fa fa fa fa
  0x0c047fff8010: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
  0x0c047fff8020: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
  0x0c047fff8030: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
  0x0c047fff8040: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
  0x0c047fff8050: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
Shadow byte legend (one shadow byte represents 8 application bytes):
  Addressable:           00
  Partially addressable: 01 02 03 04 05 06 07 
  Heap left redzone:       fa
  Freed heap region:       fd
  Stack left redzone:      f1
  Stack mid redzone:       f2
  Stack right redzone:     f3
  Stack after return:      f5
  Stack use after scope:   f8
  Global redzone:          f9
  Global init order:       f6
  Poisoned by user:        f7
  Container overflow:      fc
  Array cookie:            ac
  Intra object redzone:    bb
  ASan internal:           fe
  Left alloca redzone:     ca
  Right alloca redzone:    cb
  Shadow gap:              cc
==31==ABORTING

Answer 1

因为您的代码中没有使用 heap memory ，没有什么可能导致堆溢出，但可以解决的问题在这里:

int WordLen = sizeof(word)/sizeof(char);. 

在论证中

(char * word)   

word在作为参数传递时衰减为指针，所以

sizeof(word) 

真的会返回一个指针的大小(32 位寻址的 4 个字节)，而不是您可能传递的数组的大小。顺便说一句， sizeof(char)总是 1根据定义。
将表达式更改为:

int WordLen = strlen(word);

更改这一行可以让您的代码按我认为的那样工作。我使用以下调用方法对其进行了测试:

int main(void) 
{
    char word[] = {"this is A string"};
    bool res = detectCapitalUse(word);
    return 0;
}

一个警告 :
如果您碰巧在调用函数中使用了动态分配的内存，或者在本文未包含的其他地方使用了动态分配的内存，并且尝试错误地访问该内存块，则可能导致堆缓冲区溢出.