C-macro: set a register field defined by a bit-mask to a given value
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Published on 2010-05-26T16:47:01Z
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2010/05/26
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c
I've got 32-bit registers with field defined as bit-masks, e.g.
#define BM_TEST_FIELD 0x000F0000
I need a macro that allows me to set a field (defined by its bit-mask) of a register (defined by its address) to a given value. Here's what I came up with:
#include <stdio.h>
#include <assert.h>
typedef unsigned int u32;
/*
* Set a given field defined by a bit-mask MASK of a 32-bit register at address
* ADDR to a value VALUE.
*/
#define SET_REGISTER_FIELD(ADDR, MASK, VALUE) \
{ \
u32 mask=(MASK); u32 value=(VALUE); \
u32 mem_reg = *(volatile u32*)(ADDR); /* Get current register value */ \
assert((MASK) != 0); /* Null masks are not supported */ \
while(0 == (mask & 0x01)) /* Shift the value to the left until */ \
{ /* it aligns with the bit field */ \
mask = mask >> 1; value = value << 1; \
} \
mem_reg &= ~(MASK); /* Clear previous register field value */ \
mem_reg |= value; /* Update register field with new value */ \
*(volatile u32*)(ADDR) = mem_reg; /* Update actual register */ \
}
/* Test case */
#define BM_TEST_FIELD 0x000F0000
int main()
{
u32 reg = 0x12345678;
printf("Register before: 0x%.8X\n", reg);/* should be 0x12345678 */
SET_REGISTER_FIELD(®, BM_TEST_FIELD, 0xA);
printf("Register after: 0x%.8X\n", reg); /* should be 0x123A5678 */
return 0;
}
Is there a simpler way to do it?
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