690 - I2P2015_homework

#	Problem	Pass Rate (passed user / total user)
10432	hw2 - instruction simulator
10445	hw3 - code translation 1
10446	hw3 - code translation 2
10447	hw3 - code translation 3

10432 - hw2 - instruction simulator

Status | Limits Submit

	Time	Memory
Case 1	3 sec	16 MB
Case 2	3 sec	16 MB
Case 3	3 sec	16 MB
Case 4	3 sec	16 MB

Description

The x97 CPU has 4 32 bits registers, r0, r1, r2, and r3, whose initial values are all zeros. The CPU has 2 kinds of instructions, MOV and ADD, whose formats are specified as follows.

1. MOV , : assign constant C to register 1.

2. MOV , : copy the content of register 2 to register 1.

3. ADD , : add the value of register 1 and constant C, and save the result to register 1.

4. ADD , : add the value of register 1 and the value of register 2, and save the result to register 1.

Hint:
You can use the following code to get the input and to print the output.
#include 
#include 
#include 
enum code {MOV, ADD};
enum op_type {REG, CONST};
typedef struct INST {
    enum code opcode;
    enum op_type op1;
    int operand1_value;
    enum op_type op2;
    int operand2_value;
    struct INST* next;
} INST;
/* read input from stdin. */
INST* readInput(){
    char opc[10], op1[10], op2[10];
    int ns=0, c;
    INST *head=0, *curr=0;
    ns = scanf("%s %s %d ", opc, op1, &c);
    while(ns > 0) {
        /* allocate a storage for an instruction */
        if (head == 0){
            head = (INST*) malloc(sizeof(INST));
            curr = head;
            curr->next = 0;
        } else {
            curr->next = (INST*) malloc(sizeof(INST));
            curr = curr->next;
            curr->next = 0;
        }
       /* set the content of the instruction.*/
        /* opcode */
        if (strcmp(opc, "MOV")==0)
            curr->opcode = MOV;
        else if (strcmp(opc, "ADD")==0)
            curr->opcode = ADD;
        /* Since op1 is always a register, read its id.*/
        curr->op1 = REG;
        curr->operand1_value = op1[1]-'0';
        /* read operand 2.*/
        if (ns == 2){  /* operand 2 is a register*/
            scanf("%s ", op2);
            curr->op2 = REG;
            curr->operand2_value = op2[1] -'0';
        } else if (ns == 3) { /* operand 2 is a constant.*/
            curr->op2 = CONST;
            curr->operand2_value = c;
        }
        /* get next one*/
        ns = scanf("%s %s %d ", opc, op1, &c);
    }
    return head;
}

Input

The input is a list of instructions.

Output

The output is the values of all register in the order of r0, r1, r2, and r3.

Use %4d for each resgister and print a newline in the end.

Sample Input Download

MOV r1, 3MOV r2, 1ADD r1, r2MOV r0, 100

Sample Output Download

 100   4   1   0

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10445 - hw3 - code translation 1

Status | Limits Submit

	Time	Memory
Case 1	1 sec	32 MB

Description

We have 4 normal registers, r0, r1, r2, r3, whose initial values are all zeros, and one spe- cial register, sp, which stores stack pointer. Be careful, we cannot change the value in sp register. This CPU has 5 kinds of instructions, MOV, ADD, CMP, JMP, JLE, whose formats are specified as follows.

: Any register, r0, r1, r2, r3

: A memory address. For example, [sp+8]. Be careful, there is no space between ‘[’ and ‘]’.

: A constant value

MOV , will copy the value stored in the latter to the former

MOV ,

ADD , will add the value stored in the latter to the former

ADD ,

CMP ,

JMP

JLE if the former is smaller than the latter, jump to the label.

Please fill in the following assembly code which is a translation of the following c code using these introduced instructions and store variable “a” in r0, variable “b” in r1.

c code :

int main() {

int a, b;

a=1;

b=2;

b = a+b+1;

}

assembly code :

MOV r0, 1

MOV r1, 2

/* something missing */

ADD r1, 1

Input

There is no input.

Output

Please print the complete assembly code and use minimal instructions.

Sample Input Download

int main(){
int a=1, b=2;
a = b+3;
}You don't need to read the input. Just use your eyes.

Sample Output Download

MOV r0, 1
MOV r1, 2
MOV r0, r1
ADD r0, 3

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10446 - hw3 - code translation 2

Status | Limits Submit

	Time	Memory
Case 1	1 sec	32 MB

Description

: Any register, r0, r1, r2, r3

: A memory address. For example, [sp+8]. Be careful, there is no space between ‘[’ and ‘]’.

: A constant value

MOV , will copy the value stored in the latter to the former

MOV ,

ADD , will add the value stored in the latter to the former

ADD ,

CMP ,

JMP

JLE if the former is smaller than the latter, jump to the label.

Please fill in the following assembly code which is a translation of the following c code using these introduced instructions and store variable “a” in r0, variable “b” in r1, variable "c" in r2, variable "d" in r3.

c code :

int main(){

int a=10, b=20, c=30, d=40;

while(a<=d){

if(b<=c){

a++;

b++;

}

else if(a<=b) a++;

else a+=2;

}

assembly code :

MOV r0, 10

MOV r1, 20

MOV r2, 30

MOV r3, 40

JMP L2

L3 :

CMP r1, r2

/* something missing */

CMP r0, r1

JLE L5

ADD r0, 2

JMP L2

L5 :

/* something missing */

JMP L2

L4 :

ADD r0, 1

ADD r1, 1

L2 :

/* something missing */

Input

There is no input.

Output

Please print the complete assembly code and use minimal instructions.

Sample Input Download

int main(){
int a=1, b=2;
a = b+3;
}You don't need to read the input. Just use your eyes.

Sample Output Download

MOV r0, 1
MOV r1, 2
MOV r0, r1
ADD r0, 3

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10447 - hw3 - code translation 3

Status | Limits Submit

	Time	Memory
Case 1	1 sec	32 MB

Description

: Any register, r0, r1, r2, r3

: A memory address. For example, [sp+8]. Be careful, there is no space between ‘[’ and ‘]’.

: A constant value

MOV , will copy the value stored in the latter to the former

MOV ,

ADD , will add the value stored in the latter to the former

ADD ,

CMP ,

JMP

JLE if the former is smaller than the latter, jump to the label.

c code :

int main(){

int i,j,a,b,c,d;

a = 5;

b = 10;

c = 1;

d = 2;

for(i=0;i<=a;i++){

for(j=0;j<=b;j++){

c++;

}

d++;

}

assembly code :

MOV [sp], 5

MOV [sp+4], 10

MOV [sp+8], 1

MOV [sp+12], 2

MOV r0, 0

JMP L2

L3 :

MOV r1, 0

JMP L4

L5 :

/* something missing */

ADD r1, 1

L4 :

CMP r1, [sp+4]

/* something missing */

ADD r0, 1

L2 :

/* something missing */

JLE L3

MOV r0, [sp]

MOV r1, [sp+4]

MOV r2, [sp+8]

MOV r3, [sp+12]

Input

There is no input.

Output

Please print the complete assembly code and use minimal instructions.

Sample Input Download

int main(){
int a=1, b=2;
a = b+3;
}You don't need to read the input. Just use your eyes.

Sample Output Download

MOV r0, 1MOV r1, 2
MOV r0, r1
ADD r0, 3