C语言 使用 OpenSSL 计算并打印文件的 SHA256 哈希值
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Calculate and print SHA256 hash of a file using OpenSSL
提问by dan6470
I'm trying to write a C function using OpenSSL/libcrypto to calculate the SHA256 sum of a file. I'm basing my code on Adam Lamer's c++ example here.
我正在尝试使用 OpenSSL/libcrypto 编写一个 C 函数来计算文件的 SHA256 总和。我的代码基于 Adam Lamer 的 c++ 示例here。
Here's my code:
这是我的代码:
int main (int argc, char** argv)
{
char calc_hash[65];
calc_sha256("file.txt", calc_hash);
}
int calc_sha256 (char* path, char output[65])
{
FILE* file = fopen(path, "rb");
if(!file) return -1;
char hash[SHA256_DIGEST_LENGTH];
SHA256_CTX sha256;
SHA256_Init(&sha256);
const int bufSize = 32768;
char* buffer = malloc(bufSize);
int bytesRead = 0;
if(!buffer) return -1;
while((bytesRead = fread(buffer, 1, bufSize, file)))
{
SHA256_Update(&sha256, buffer, bytesRead);
}
SHA256_Final(hash, &sha256);
sha256_hash_string(hash, output);
fclose(file);
free(buffer);
return 0;
}
void sha256_hash_string (char hash[SHA256_DIGEST_LENGTH], char outputBuffer[65])
{
int i = 0;
for(i = 0; i < SHA256_DIGEST_LENGTH; i++)
{
sprintf(outputBuffer + (i * 2), "%02x", hash[i]);
}
outputBuffer[64] = 0;
}
The problem is this....take a look at the calculated sums below for an example file:
问题是这样的....看看下面的计算总和的示例文件:
Known good SHA256: 6da032d0f859191f3ec46a89860694c61e65460d54f2f6760b033fa416b73866
Calc. by my code: 6dff32ffff59191f3eff6affff06ffff1e65460d54ffff760b033fff16ff3866
I also get * stack smashing detected * when the code is finished executing.
当代码完成执行时,我还检测到 * stack smashing *。
Does anyone see what I'm doing wrong?
有没有人看到我做错了什么?
Thanks!
谢谢!
回答by P.T.
Looks like there are a lot of '0xff' blocks in your output, and the corresponding blocks in the good string have the high bit set ... maybe a sign extension problem somewhere.
看起来您的输出中有很多 '0xff' 块,并且好字符串中的相应块设置了高位......也许某处的符号扩展问题。
Does making:
是否制作:
char hash[SHA256_DIGEST_LENGTH];
unsigned, like:
未签名,例如:
unsigned char hash[SHA256_DIGEST_LENGTH];
help? (Especially in the signature of sha256_hash_string.)
帮助?(特别是在 . 的签名中sha256_hash_string)
回答by Adam Rosenfield
You're printing out a signedcharas an integer. If the byte is negative, it get's converted to a signed int(the default argument promotionsin the call to sprintf), and then that gets converted to an unsigned int(via the %xformat specifier) and printed out.
您正在打印一个带符号char的整数。如果字节为负数,则将其转换为 a signed int(调用中的默认参数提升sprintf),然后将其转换为 an unsigned int(通过%x格式说明符)并打印出来。
So, the byte A0is -96 as a signed byte, which gets converted to -96 as a signed int, which is 0xFFFFFFA0 in hex, so it gets printed out as FFFFFFA0.
因此,该字节A0是 -96 作为有符号字节,它被转换为 -96 作为 a signed int,在十六进制中是 0xFFFFFFA0,因此它被打印为 FFFFFFA0。
To fix this, case each byte to an unsigned charbefore printing:
要解决此问题,请unsigned char在打印前将每个字节设置为一个:
sprintf(..., (unsigned char)hash[i]);
You're getting the warning about stack smashing because there's a signed byte near the end of the hash, so you're writing the 8 bytes FFFFFFB7 at offset 58 when you intended to only write 2 bytes. This results in a buffer overflow, which happens to be detected here because the compiler likely inserted a guard area or security cookie in the stack before the return value, and it detected that that guard area was inadvertently modified.
您收到有关堆栈粉碎的警告,因为在散列末尾附近有一个带符号的字节,因此当您打算仅写入 2 个字节时,您正在偏移 58 处写入 8 个字节的 FFFFFFB7。这会导致缓冲区溢出,这恰好在这里被检测到,因为编译器可能在返回值之前在堆栈中插入了一个防护区或安全 cookie,并且它检测到该防护区被无意中修改了。

