如果传递给sscanf的参数被强制转换会发生什么

Question

While reviewing and old piece of code, I stumbled upon some coding horror like this one:

struct Foo
{
    unsigned int  bar;
    unsigned char qux;
    unsigned char xyz;
    unsigned int  etc;
};

void horror(const char* s1, const char* s2, const char* s3, const char* s4, Foo* foo)
{
    sscanf(s1, "%u", &(foo->bar));
    sscanf(s2, "%u", (unsigned int*) &(foo->qux));
    sscanf(s3, "%u", (unsigned int*) &(foo->xyz));
    sscanf(s4, "%u", &(foo->etc));
}

So, what is actually happening in the second and third sscanf , with the argument passed being a unsigned char* cast to unsigned int* , but with the format specifier for an unsigned integer? Whatever happens is due to undefined behavior, but why is this even "working"?

As far as I know, the cast effectively does nothing in this case (the actual type of the arguments passed as ... is unknown to the called function). However this has been in production for years and it has never crashed and the surrounding values apparently are not overwritten, I suppose because the members of the structure are all aligned to 32 bits. This is even reading the correct value on the target machine (a little endian 32 bit ARM) but I think that it would no longer work on a different endianness.

Bonus question: what is the cleanest correct way to do this? I know that now we have the %hhu format specifier (apparently introduced by C++11), but what about a legacy C89 compiler?

---

Please note that the original question had uint32_t instead of unsigned int and unsigned char instead of uint8_t but that was just misleading and out of topic, and by the way the original code I was reviewing uses its own typedefs.

Answer 1

In this case from the pointer point of view nothing as on the all modern machines pointers are the same for all types.

But because you use wrong formats - the scanf will write outside the memory allocated to the variables and it is an Undefined Behaviour.

Answer 2

Bonus question: what is the cleanest correct way to do this? I know that now we have the %hhu format specifier (apparently introduced by C++11), but what about a legacy C89 compiler?

The <stdint.h> header and its types were introduced in C99, so a C89 compiler won't support them except as an extension.

The correct way to use the *scanf() and *printf() families of functions with the various fixed or minimum-width types is to use the macros from <inttypes.h> . For example:

#include <inttypes.h>
#include <stdlib.h>
#include <stdio.h>

int main(void) {
  int8_t foo;
  uint_least16_t bar;

  puts("Enter two numbers");
  if (scanf("%" SCNd8 " %" SCNuLEAST16, &foo, &bar) != 2) {
    fputs("Input failed!\n", stderr);
    return EXIT_FAILURE;
  }
  printf("You entered %" PRId8 " and %" PRIuLEAST16 "\n", foo, bar);
}

Answer 3

First of all, this of course invokes Undefined Behaviour.

But that kind of horror was quite common in old code, where the C language was used as a higher level assembly language. So here are 2 possible behaviours:

• the structure has a 32 bits alignment. All is (rather fine) on a little endian machine: the uint8_t members will recieve the least significant byte of the 32 bits value and the padding bytes will be zeroed (I assume that the program does not try to store a value greater than 255 into an uint8_t )
• the structure has not a 32 bits alignement, but the architecture allows scanf to write into mis-aligned variables. The least significant byte of the value read for qux will correctly go into qux and the next 3 zero bytes will erase xyz and etc . On next line, xyz receives its value and etc recieves one more 0 byte. And finally etc will recieve its value. This could have been a rather common hack in the early 80' on an 8086 type machine.

For a portable way, I would use an temporary unsigned integer:

uint32_t u;
sscanf(s1, "%u", &(foo->bar));
sscanf(s2, "%u", &u);
foo->qux = (uint8_t) u;
sscanf(s3, "%u", &u);
foo->xyz = (uint8_t) u;
sscanf(s4, "%u", &(foo->etc));

and trust the compiler to generate code as efficient as the horror way.

Answer 4

OP code is UB as scan specifiers does not match arguments.

cleanest correct way to do this?

Cleaner

#include <inttypes.h>

void horror1(const char* s1, const char* s2, const char* s3, const char* s4, Foo* foo) {
    sscanf(s1, "%" SCNu32, &(foo->bar));
    sscanf(s2, "%" SCNu8, &(foo->qux));
    sscanf(s2, "%" SCNu8, &(foo->xyz));
    sscanf(s1, "%" SCNu32, &(foo->etc));
}

Cleanest

Add additional error handling if desired.

void horror2(const char* s1, const char* s2, const char* s3, const char* s4, Foo* foo) {
    foo->bar = (uint32_t) strtoul(s1, 0, 10);
    foo->qux = (uint8_t) strtoul(s1, 0, 10);
    foo->xyz = (uint8_t) strtoul(s1, 0, 10);
    foo->etc = (uint32_t) strtoul(s1, 0, 10);
}