强制转换指针时崩溃-C

Question

C-Beginner Question:

I have the following code that crashes at line unsigned long rdwValue = *((unsigned long*)pParamPtr); . I am cross compiling the code for a MIPS32 target on a fedora machine using GCC

unsigned char* pParamPtr = GetNvParamRamAddress(ParameterId, swIndex);

if (0 != pParamPtr)
{
    unsigned long rdwValue = *((unsigned long*)pParamPtr);// CRASHES
}

But if I change the line inside the if to

rdwValue = *(pParamPtr);

this works.

Am I doing something against the rules with this typecasting?

What I need is to get a four bytes [unsigned long is 4 bytes] from address starting at pParamPtr into rdwValue.

Is it memcpy the way to go?

Answer 1

What platform is this running on?

One likely reason is alignment restrictions being violated, if the returned pointer is odd and the platform doesn' support reading unsigned long s from odd addresses this would fail.

UPDATE According to a comment the OP is running this on MIPS, which certainly has alignment restrictions.

As a minor stylistic point, your outer pair of parenthesis aren't needed, it could be just:

unsigned long rdwValue = *(unsigned long *) pParamPtr;

It might be easiest to read the bytes one by one, that also gives you a chance to handle endianness explicitly. Assuming little-endian (and at least 32-bits of int precision):

unsigned long rwdValue = pParamPtr[0] | (pParamPtr[1] << 8) | (pParamPtr[2] << 16) | (pParamPtr[3] << 24);