将C递归阶乘函数转为MIPS汇编时遇汇编错误求助

Let's walk through fixing those two errors in your MIPS assembly code—they're both common pitfalls when translating C to MIPS, so you're not alone!

Error 1: .ascii directive cannot appear in text segment #

You correctly placed your string in the .rdata (read-only data) segment, but two issues are causing this error:

• You used Chinese quotation marks（“”）instead of standard English quotation marks (""). The assembler can't parse Chinese quotes, so it misinterprets the string definition as invalid text-segment code.
• While manually adding \000 works, using .asciiz is cleaner—it automatically appends the null terminator required for C-style strings (like the one printf expects).

Error 2: "$L1": operand is of incorrect type #

The jr instruction is designed to jump to an address stored in a register (e.g., jr $ra to return to the caller). Since $L1 is a direct label (a fixed memory address), you need to use the j instruction instead, which jumps directly to a label.

Corrected Full MIPS Code #

.text
.globl main
main:
    subu $sp,$sp,32 # Stack frame is 32 bytes long
    sw $ra,20($sp) # Save return address
    sw $fp,16($sp) # Save old frame pointer
    addiu $fp,$sp,28 # Set up frame pointer
    li $a0,10 # Put argument (10) in $a0
    jal fact # Call factorial function
    la $a0,$LC # Put format string in $a0
    move $a1,$v0 # Move fact result to $a1
    jal printf # Call the print function
    lw $ra,20($sp) # Restore return address
    lw $fp,16($sp) # Restore frame pointer
    addiu $sp,$sp,32 # Pop stack frame
    jr $ra # Return to caller

.rdata
$LC: .asciiz "The factorial of 10 is %d\n"

.text
fact:
    subu $sp,$sp,32 # Stack frame is 32 bytes long
    sw $ra,20($sp) # Save return address
    sw $fp,16($sp) # Save frame pointer
    addiu $fp,$sp,28 # Set up frame pointer
    sw $a0,0($fp) # Save argument (n) to use for Recursive Call
    lw $v0,0($fp) # Load n
    bgtz $v0,$L2 # Branch if n > 0
    li $v0,1 # Return 1
    j $L1 # Jump to code to return (fixed from jr to j)
$L2:
    lw $v1,0($fp) # Load n
    subu $v0,$v1,1 # Compute n - 1
    move $a0,$v0 # Move value to $a0
    jal fact # Call factorial function
    lw $v1,0($fp) # Load n
    mul $v0,$v0,$v1 # Compute fact(n-1) * n
$L1: # Result is in $v0
    lw $ra, 20($sp) # Restore $ra
    lw $fp, 16($sp) # Restore $fp
    addiu $sp, $sp, 32 # Pop stack
    jr $ra # Return to caller

This code should compile and run correctly, outputting the factorial of 10 as expected.

内容的提问来源于stack exchange，提问作者Fathima Reeza