问题描述
我想在我的C程序中编写一个非常简单的内联汇编例程，该例程不执行任何操作否则，将本地寄存器％l0-％l7设置为不同的值.我尝试了以下简单的方法:
I wanna write a very simple inline assembly routine in my C program which does nothingelse then setting the local registers %l0 - %l7 to different values. I tried the following straightforward approach:
asm volatile (
".text\n\t"
"mov 0, %%l0 \n\t"
"mov 1, %%l1 \n\t"
"mov 2, %%l2 \n\t"
"mov 3, %%l3 \n\t"
"mov 4, %%l4 \n\t"
"mov 5, %%l5 \n\t"
"mov 6, %%l6 \n\t"
"mov 7, %%l7 \n\t"
);
不幸的是，汇编器告诉:每条指令的操作数非法.有人可以这么高兴向我指出如何正确地将立即值传递给SPARC汇编程序吗?
unfortuantely, the assembler tells: illegal operand for each instruction. Could somebody please be so nice to point me out how I can properly pass immediate values to the SPARC assembler?
非常感谢！
谢谢克里斯，我进行了您建议的更改，但Sparc编译器仍然告诉您一些有关非法操作数的信息...
Thanks Chris, I made the changes you suggested but the Sparc compiler still tells some something about illegal operands...
推荐答案
SPARC并不具有这样的立即移动"指令；有一个or可以像or %g0, 123, %l0一样使用(或者使用零寄存器%g0使用不超过11位的常数，从而将该常数移到目标寄存器中)，或者sethi指令，可用于设置寄存器的 upper 21位.为了容纳 any (32位)常量，因此必须合成两步式set，首先对高位进行sethi，然后对低位进行or.
SPARC doesn't mave "immediate move" instructions as such; there's either or which can be used like or %g0, 123, %l0 (or'ing a no-more-than 11-bit constant with the zero register, %g0, resulting in moving said constant into the target register), or the sethi instruction that can be used to set the upper 21 bits of a register. In order to accommodate any (32bit) constant, you therefore have to synthesise a two-step set by doing a sethi first for the upper bits followed by an or with the lower ones.
SPARC汇编程序通常知道创建该序列的set ..., %tgtregister快捷方式，和/或在常数适合的情况下消除一条指令.
SPARC assemblers usually know a set ..., %tgtregister shortcut to create this sequence, and/or eliminate one instruction should the constant be fit for that.
还请注意，在64位/sparcv9中，set指令最终可能会评估为最多5条指令的序列，从而将事物移位/或移位.
Also note that in 64bit / sparcv9, the set instruction may ultimately evaluate to a sequence of up to five instructions, shifting/or'ing things together.
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