Question

如果我编码

    asm volatile("" ::: "r1", "r2", "r3", "r4", "r5", "r6", "r7", "r8",
                    "r10", "r11", "r12", "r13", "r14", "r15", "r16", "r17",
                    "r18", "r19", "r20", "r21", "r22", "r23", "r24", "r25",
                    "r26", "r27", "r28", "r29", "r30", "r31", "memory");

并且反汇编avr-gcc的结果我可以看到avr-gcc假定X，Y和Z没有被破坏。然而，它们被隐含地声明为破坏，因为它们是对（r26，r27），（r28，r29）和（r30，r31）。 avr-gcc似乎将X，Y，Z视为单独的寄存器。

如果我试图解决这个问题

    asm volatile("" ::: "r1", "r2", "r3", "r4", "r5", "r6", "r7", "r8",
                    "r10", "r11", "r12", "r13", "r14", "r15", "r16", "r17",
                    "r18", "r19", "r20", "r21", "r22", "r23", "r24", "r25",
                    "r26", "r27", "r28", "r29", "r30", "r31", "X", "Y", "Z", "memory");

然后avr-gcc抱怨

test.cpp:212:91: error: unknown register name ‘Z’ in ‘asm’
test.cpp:212:91: error: unknown register name ‘Y’ in ‘asm’
test.cpp:212:91: error: unknown register name ‘X’ in ‘asm’

根据编译器错误以及根据Register Layout的文档，看起来好像无法指定X，Y和Z.

我怎样才能告诉avr-gcc除“r0”之外的所有寄存器都被破坏了？特别是如何判断X，Y，Z是否被破坏？

Clobber X，Y，Z使用avr-gcc注册

0 个答案: