静态代码分析工具报告的缓冲区溢出问题

Question

My piece of code:

void temp(char *source)
{
    char dest[41];

    for(int i = 0; i < 20; i++)
    {
        sprintf(&dest[i*2], "%02x", (unsigned int)source[i]);
    }
}

When I run the static code analysis tool, I get the warning below:

On 19th iteration of the loop : This code could write past the end of the buffer pointed to by &dest[i * 2]. &dest[i * 2] evaluates to [dest + 38]. sprintf() writes up to 9 bytes starting at offset 38 from the beginning of the buffer pointed to by &dest[i * 2], whose capacity is 41 bytes.The number of bytes written could exceed the number of allocated bytes beyond that offset. The overrun occurs in stack memory.

My question is: since in every loop iteration, we are only copying 2 bytes (considering size of unsigned int on the machine is 2 bytes) from source to destination, where is the possibility of copying 9 bytes on the last iteration?

Answer 1

char can be signed, and is so by default in x86 compilers. On my computer

#include <stdio.h>

int main(void) {
    printf("%02x\n", (unsigned int)(char)128);
}

prints ffffff80 .

What you want to do is use format "%02hhx" and argument (unsigned char)c .