CSc 256 Midterm 1 Fall 2011

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1 CSc 256 Midterm 1 Fall 2011 NAME: Problem 1a: Given the C++ function prototype and variable declarations: int func(int arg0, int *arg1, int *arg2); int *ptr, n, arr[10]; which of the following statements will cause syntax errors or warnings about type mismatch/missing cast? For each case, write either BAD (error or warning), or OK (no error or warning), and circle also *all* the argument/variables that have type mismatches. No points will be awarded for a statement if there is a syntax error and you circle the wrong argument/variable. (6 points) *ptr = func(ptr[n], &arr[n], ptr+1); OK arr[n] = func(arr[*ptr], &arr[n], &ptr); BAD; &ptr is not a pointer to an int

2 Problem 1b: Suppose we are tracing a C/C++ program using a debugger such as gdb. The code showing all function calls looks like this: main() { F1(- 1); F1(3); F1(int arg) { F2(arg+2); F2(arg/2); F2(int arg) { F3(arg); F3(arg*2); int x=4; // some code not shown F3(int arg) { // some code not shown Suppose we set a breakpoint at the line int x=4 (in F2()). We run the program, stop at the breakpoint for the first time, and check the stack frames on the stack using a command similar to gdb s backtrace. What are the stack frames that you see at this time? For each stack frame that you see, indicate clear what function it belongs to, the arguments for that function call, and the order (which one is at the bottom, which is above it, etc). (4 points)

3 Problem 2: Translate the C++ function upt() into MIPS assembly language. (You don't have to write a main program, or the dst() function.) Your solution must compile and run correctly in spim, and follow all MIPS register use conventions. You may use $s? or $t? as temporaries, but follow the specifications below. For upt: void upt(int arg0, int *arg1, int *arg2) arg0 a0 arg1 a1 arg2 a2 For dst: int dst(int arg0, int *arg1) arg0 a0 arg1 a1 result returned in $v0 Make sure your loops are efficient (i.e., they should not have unnecessary branches). Points will be deducted for obvious inefficiencies. (45 points)

4 void upt(int arg0, int *arg1, int *arg2) { int *ptr, prod=0; ptr = arg2; for (int i=arg0; i>=0; i=i/2) { ptr++; if (arg0 < *ptr prod > 2) { arg1[1] = arg1[1] *2; else { prod--; prod = prod * dst(arg1[i], ptr);

5 # $s0 copy of arg0 # $s1 copy of arg1 # $s2 ptr # $s3 prod # $s4 i upt: sw $ra, -4($sp) sw $s0, -8($sp) sw $s1, -12($sp) sw $s2, -16($sp) sw $s3, -20($sp) sw $s4, -24($sp) addi $sp, $sp, -24 move $s0, $a0 move $s1, $a1 li $s3, 0 # int *ptr, prod=0; move $s2, $a2 # ptr = arg2; move $s3, $s0 # for (int i=arg0; i>=0; i=i/2) { blt $s4, $0, exit for: addi $s2, $s2, 4 # ptr++; lw $t0, ($s2) # if (arg0 < *ptr prod > 2){ blt $s0, $t0, if ble $s3, 2, else if: lw $t0, 4($s1) # arg1[1] = arg1[1] *2; mul $t0, $t0, 2 sw $t0, 4($s1) # j call else: # else { addi $s3, $s3, -1 # prod--; call: # sll $t0, $s4, 2 # prod = prod * add $t0, $t0, $s1 # dst(arg1[i], ptr); lw $a0, ($t0) move $a1, $s2 jal dst mul $s3, $s3, $v0 div $s4, $s4, 2 bge $s4, $0, for exit:addi $sp, $sp, 24 lw $ra, -4($sp) lw $s0, -8($sp) lw $s1, -12($sp) lw $s2, -16($sp) lw $s3, -20($sp) lw $s4, -24($sp) jr $ra

6 Problem 3: Show what is printed on the screen after this C++ program is executed. (20 points) ANSWER: occupawaslst #include <iostream> using std::cout; int main() { char str[] = "occupywallst"; char *ptr0, *ptr1; ptr0 = &str[8]; ptr1 = ptr0-5; *ptr0 = (*ptr1) - 2; ptr0 = ptr0-1; ptr1 = ptr1 + 2; *ptr1 = *ptr0; cout << str; [Work:] ptr1 = ptr0-5; *ptr0 = (*ptr1) - 2; ptr0 = ptr0-1; ptr1 = ptr1 + 2; *ptr1 = *ptr0; ptr1 = &str[3] str[8] = str[3] 2 = s ptr0 = &str[7] ptr1 = &str[5] str[5] = str[7] = a

7 Problem 4: The following MIPS program is executed. Show the contents of registers $t0, $t1 and $t2, when the label "end" is reached. Assume the initial contents of registers as shown. All integers are in hex. Show each step clearly for partial credit. (25 points) Initially: $t0 0x $t1 0x $t2 0xffffff9c label address contents here: 0x xc5c6c7c8 there: 0x x text main:srl $t2,$t2,5 sb $t2,-2($t0) or $t0, $t2, 0x lw $t1,here end: li syscall $v0,10 CONTENTS of $t0 (in hex): 0x07fffffe CONTENTS of $t1 (in hex): 0xc5c6fcc8 CONTENTS of $t2 (in hex): 0x07ffffc $t2 = srl $t2,$t2,5 $t2 = = 0x07ff fffc sb $t2,-2($t0) or byte at 0x = 0xfc (replaces 0xc7) $t0, $t2, 0x $t2 = $t0 = = 0x07fffffe lw $t1, here $t1 = 0xc5c6fcc8

8 MIPS instructions op1, op2 are registers, op3 is register or constant cont[op1] means contents of op1 move op1, op2 cont[op1] = cont[op2] add op1, op2, op3 cont[op1] = cont[op2] + cont[op3] sub op1, op2, op3 cont[op1] = cont[op2] - cont[op3] mul op1, op2, op3 cont[op1] = cont[op2] * cont[op3] div op1, op2, op3 cont[op1] = cont[op2] / cont[op3] rem op1, op2, op3 cont[op1] = cont[op2] % cont[op3] not op1, op2 cont[op1] = not cont[op2] (bitwise) and op1, op2, op3 cont[op1] = cont[op2] and cont[op3] (bitwise) or op1, op2, op3 cont[op1] = cont[op2] or cont[op3] (bitwise) nand op1, op2, op3 cont[op1] = cont[op2] nand cont[op3] (bitwise) nor op1, op2, op3 cont[op1] = cont[op2] nor cont[op3] (bitwise) xor op1, op2, op3 cont[op1] = cont[op2] xor cont[op3] (bitwise) sll op1, op2, AMT cont[op1] = cont[op2] shift left logical by AMT bits srl op1, op2, AMT cont[op1] = cont[op2] shift right logical by AMT bits sra op1, op2, AMT cont[op1] = cont[op2] shift right arithmetic by AMT bits rol op1, op2, AMT cont[op1] = cont[op2] rotate left by AMT bits ror op1, op2, AMT cont[op1] = cont[op2] rotate right by AMT bits b label j label beq op1, op2, label bne op1, op2, label bgt op1, op2, label bge op1, op2, label blt op1, op2, label ble op1, op2, label beqz op1, label bnez op1, label bgtz op1, label bgez op1, label bltz op1, label blez op1, label goto label goto label if (cont[op1]==cont[op2]) goto label if (cont[op1]!=cont[op2]) goto label if (cont[op1]>cont[op2]) goto label if (cont[op1]>=cont[op2]) goto label if (cont[op1]<cont[op2]) goto label if (cont[op1]<=cont[op2]) goto label if (cont[op1]==0) goto label if (cont[op1]!=0) goto label if (cont[op1]>0) goto label if (cont[op1]>=0) goto label if (cont[op1]<0) goto label if (cont[op1]<=0) goto label la R, label li R, constant lw R,?? lb R,?? cont[r] = address of label cont[r] = constant cont[r] = M[ADDR] cont[r] = m[addr], sign- extended

9 lbu R,?? sw R,?? sb R,?? cont[r] = m[addr], zero- extended M[ADDR] = cont[r] m[addr] = low 8- bits of cont[r] if?? is a label, ADDR = address of label if?? is (R), ADDR = cont[r] if?? is constant(r), ADDR = cont[r] + constant if?? is label(r), ADDR = cont[r] + address of label mtc0 op1, op2 contents of coprocessor 0 register op1 = contents of MIPS register op2 mfc0 op1, op2 contents of MIPS register op1 = contents of coprocessor 0 register op2 Syscall usage: print an int $v0=1, $a0=int to be printed print a string $v0=4, $a0=address of string to be printed read an int $v0=5, input int appears in $v0 exit $v0=10 MIPS register names: $0 $1 $2,$3 $v0,$v1 $4 - $7 $a0 - $a3 $8 - $15 $t0 - $t7 $16 - $23 $s0 - $s7 $24 - $25 $t8 - $t9 $26 - $27 $k0 - $k1 $28 $gp $29 $sp $30 $s8 $31 $ra

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