Question

This is my code:

volatile uint32_t value = *((volatile uint32_t *) 0xA0000000); // here `value` is 12498
value *= 2; // here `value` is still 12498
value |= 0x0000001; // still 12498

When analysing the variable value in my debugger, it holds the same value on all lines. What am I doing wrong?

Answer 1

Maybe your debugger isn’t actually that good. When one of my tools doesn’t seem to behave, I always check it against another tool.

Try to debug the old-fashioned way, with:

volatile uint32_t value = *((volatile uint32_t *) 0xA0000000);
printf ("A:%d\n", value);

value *= 2;
printf ("B:%d\n", value);

value |= 0x0000001;
printf ("C:%d\n", value);

or some other output method if printf is unavailable (it looks like you may be working in the embedded space).

See what you get with that – I’d be more inclined to trust printf-debugging than debuggers.

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If your problem is not with value but instead with the memory at location 0xA0000000, then it’s working as expected.

You’re manipulating the local variable, not the memory location. You need to write the value back, with something like:

*((volatile uint32_t *) 0xA0000000) = value;

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However, given your use of volatile, it’s entirely possible you just wanted a variable pointer to that location so that changes would reflect immediately.

If that’s the case, you would need something along the lines of:

volatile uint32_t *pValue = (volatile uint32_t *) 0xA0000000;
*pValue *= 2;
*pValue |= 0x00000001;

In that case, the memory location would be changed at each instruction without having to explicitly write the value.