因此,不是: *cc = *cc + 100; mov r3, #0 ldr r1, [r3, #0] add r1, r1, #100 str r1, [r3, #0] 它有没有类似或接近的东西: add [r3, #0],#100 ;I know this is wrong in current arm instruction set 浏览0提问于2015-08-06得票数 0 3回答 arm-none-eabi-as和ARM评论标志‘...
ARM LDR伪指令使用方法具体解释 LDR伪指令 10.45 LDR pseudo-instruction 功能:把一个32位马上数或一个32位的内存地址载入到一个寄存器中。 注意:这里描写叙述的是LDR伪指令,而不是LDR指令 语法: LDR{cond}{.W} Rt, =expr LDR{cond}{.W} Rt, =label_expr *cond是一个可选的条件码 *.W是可选的指令宽...
ARM LDR伪指令使用方法具体解释 LDR伪指令 10.45 LDR pseudo-instruction 功能:把一个32位马上数或一个32位的内存地址载入到一个寄存器中。 注意:这里描写叙述的是LDR伪指令,而不是LDR指令 语法: LDR{cond}{.W} Rt, =expr LDR{cond}{.W} Rt, =label_expr *cond是一个可选的条件码 *.W是可选的指令宽...
From arm arm. The label of the literal data item that is to be loaded into <Rt>. The assembler calculates the required value of the offset from the PC value of this instruction to the label. Permitted values of the offset are: Encoding T1 multiples of four in the range 0 to 1020 ...
This pseudo op will always evaluate to a legal ARM instruction that does nothing. Currently it will evaluate to MOV r0, r0. LDR ldr <register> , = <expression> If expression evaluates to a numeric constant then a MOV or MVN instruction will be used in place of the LDR instruction, if...
after that I fail.For example the r2 come out 0x05 and at the end r0 come out 0x10C how can this be,there is no code here to save the r0 on to the memory why it changes.I guess my dont know what to do with r2 and r4.Anyone has knowlage about ARM help will be really good...
subtract 8 from the offset value to allow for ARM920T pipelining. In this case base write-back should not be specified. <Address>can be:(以下是关于地址的几种表示方法) 1 An expression which generates an address:(立即寻址) The assembler will attempt to generate an instruction using the PC ...
#0参数是使用mov,不是ldr 解决:改为mov...built-in function 'strcpy' 错误:strcpy函数与内置函数库冲突,(比如:printf,strlen,puts,scanf等都是内置函数库) 解决: 1.在arm-linux-gcc...:1.在本文开头上添加: extern void uart0_init(void); 2.添加该函数的头文件,例如: #include "serial.h" arm-...
Example 12 shows an ARM code routine that overwrites one string with another string. It uses the LDR pseudo-instruction to load the addresses of the two strings from a data section. The following are particularly significant: DCB The DCB directive defines one or more bytes of store. In addit...
The behavior you are seeing is indeed strange. Can you open "Tools"->"ARM Advanced Features" window and check if the MMU and/or any of the caches are enabled ? If the cache is enabled, it is possible that the "ldr" instruction is fetching from the cache and not RAM...