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How an assembler converts assembly mnemonics to machine language?
Each mnemonic maps directly to a machine instruction code, known as an opcode. Some mnemonics map to more than one opcode, however the instruction's operand types will determine which specific opcode will be generated.
What is the need of two pass assembler?
#include<stdio.h> #include<conio.h> #include<string.h> struct symbol { char sn[35]; int add; }s[35]; void main() { int symerr=0,oper=0,i=1,j=1,lc,k,sa,p1,f,op,l,q,d,x=1; char opcode[50],lb[25],operand[25]; char optab[10][10]={"LDA","LDB","STA","ADD","COMP","J","JEQ","SUB","STA"}; char value[10][10]={"00","b8","0c","18","28","3C","30","lc","14","78"}; FILE *fp1,*fp2; clrscr(); fp1=fopen("input1.txt","r"); fp2=fopen("inter.txt","w"); fscanf(fp1,"%s %s",&lb,&opcode); if(strcmp(opcode,"START")==0) { fscanf(fp1,"%d",&lc); sa=lc; fprintf(fp2,"%d\t %s\t %s\t %d\n",lc,lb,opcode,lc); fscanf(fp1,"%s %s",&lb,&opcode); } while(!feof(fp1)) { d=lc; if(strcmp(".",lb)!=0) { if(strcmp("-",lb)!=0) { for(i=1;i<=j;i++) { if(strcmp(s[i].sn,lb)==0) symerr=1; break; } if(symerr==0) { strcpy(s[j].sn,lb); s[j].add=lc; j++; } } for(k=0;k<10;k++) for(l=0;l<10;l++) if(optab[k][l]==opcode) { fscanf(fp1,"%s",&operand); lc=lc+3; x=0; } if(strcmp(opcode,"RESW")==0) { fscanf(fp1,"%s",&operand); op=atoi(operand); lc=lc+(op*3); } else if(strcmp(opcode,"RESB")==0) { fscanf(fp1,"%s",&operand); op=atoi(operand); lc=lc+op; } else if(strcmp(opcode,"BYTE")==0) { fscanf(fp1,"%s",&operand); f=strlen(operand); lc=lc+(f-3); } else if(strcmp(opcode,"WORD")==0) { fscanf(fp1,"%s",&operand); lc=lc+3; } else { if(x==1) { fscanf(fp1,"%s",operand); lc=lc+3; } } } fprintf(fp2,"\n%d\t%s\t%s\t%s\n",d,lb,opcode,operand); if(symerr==1) fprintf(fp2,"\n**DUPLICATE SYMBOL**\n"); if(oper==1) fprintf(fp2,"\n**INVALID OPERATION CODE**\n"); fscanf(fp1,"%s\t%s",&lb,&opcode); symerr=0; oper=0; } p1=lc-sa; fprintf(fp2,"\nThe program length is %d",p1); printf("\n symbol table\tlabeL address\n"); for(q=1;q<j;q++) printf("\n%s\t%d",s[q].sn,s[q].add); fcloseall(); getch(); }
SIC assembler pass 1 C program?
#include<stdio.h> #include<string.h> #include<stdlib.h> void chk_label(); void chk_opcode(); void READ_LINE(); struct optab { char code[10],objcode[10]; }myoptab[3]={ {"LDA","00"}, {"JMP","01"}, {"STA","02"} }; struct symtab{ char symbol[10]; int addr; }mysymtab[10]; int startaddr,locctr,symcount=0,length; char line[20],label[8],opcode[8],operand[8],programname[10]; void PASS1() { FILE *input,*inter; input=fopen("input.txt","r"); inter=fopen("inter.txt","w"); printf("LOCATION LABEL\tOPERAND\tOPCODE\n"); printf("_____________________________________"); fgets(line,20,input); READ_LINE(); if(!strcmp(opcode,"START")) { startaddr=atoi(operand); locctr=startaddr; strcpy(programname,label); fprintf(inter,"%s",line); fgets(line,20,input); } else { programname[0]='\0'; startaddr=0; locctr=0; } printf("\n %d\t %s\t%s\t %s",locctr,label,opcode,operand); while(strcmp(line,"END")!=0) { READ_LINE(); printf("\n %d\t %s \t%s\t %s",locctr,label,opcode,operand); if(label[0]!='\0') chk_label(); chk_opcode(); fprintf(inter,"%s %s %s\n",label,opcode,operand); fgets(line,20,input); } printf("\n %d\t\t%s",locctr,line); fprintf(inter,"%s",line); fclose(inter); fclose(input); } void READ_LINE() { char buff[8],word1[8],word2[8],word3[8]; int i,j=0,count=0; label[0]=opcode[0]=operand[0]=word1[0]=word2[0]=word3[0]='\0'; for(i=0;line[i]!='\0';i++) { if(line[i]!=' ') buff[j++]=line[i]; else { buff[j]='\0'; strcpy(word3,word2); strcpy(word2,word1); strcpy(word1,buff); j=0; count++; } } buff[j-1]='\0'; strcpy(word3,word2); strcpy(word2,word1); strcpy(word1,buff); switch(count) { case 0:strcpy(opcode,word1); break; case 1:strcpy(opcode,word2); strcpy(operand,word1); break; case 2:strcpy(label,word3); strcpy(opcode,word2); strcpy(operand,word1); break; } } void chk_label() { int k,dupsym=0; for(k=0;k<symcount;k++) if(!strcmp(label,mysymtab[k].symbol)) { mysymtab[k].addr=-1; dupsym=1; break; } if(!dupsym) { strcpy(mysymtab[symcount].symbol,label); mysymtab[symcount++].addr=locctr; } } void chk_opcode() { int k=0,found=0; for(k=0;k<3;k++) if(!strcmp(opcode,myoptab[k].code)) { locctr+=3; found=1; break; } if(!found) { if(!strcmp( opcode,"WORD")) locctr+=3; else if (!strcmp(opcode,"RESW")) locctr+=(3*atoi(operand)); else if(!strcmp(opcode,"RESB")) locctr+=atoi(operand); } } int main() { PASS1(); length=locctr-startaddr; }
What is the difference between prefix and postfix increment operator in c plus plus?
Both the prefix and the postfix increment operators increment the operand. The difference is what is the value of the expression during the evaluation of the expression. In the prefix form, the value is already incremented. In the postfix form, it is not. int a = 1; int b = ++a; // both a and b are now equal to 2 int a = 1; int b = a++; // a is equal to 2 and b is equal to 1
How do you convert opcode to machine code?
popfd
What is the difference between operand and opcode?
op code is used as the value of instruction . And operand is address location where the instruction can meet.
What is operand difference between operand and operator?
Simply defining, in an expression like A+B A is an Operand B is an Operand Plus is the Operator in between
How does 8086 differentiate between an opcode and operand?
The 8086 microprocessor differentiates between an opcode and an operand primarily through the instruction format, where the opcode is always specified first, followed by the operands. The opcode indicates the operation to be performed, while the operands represent the data or addresses on which the operation will act. The instruction's length is variable, and the processor uses specific bits in the instruction to determine the types and sizes of operands, allowing it to interpret the instruction correctly. Additionally, the opcode itself can include information about the addressing mode, further aiding in the distinction between opcodes and operands.
Specify the opcode and operand in the instruction MOV HL?
mov H , L mov is opcode H L are operands
What are the Machine Cycles For Inr M Instruction?
3 for opcode fetch, 1 for opcode decode, 3 for operand fetch, and 3 for opcode store, for a total of 10, not including wait states.
What do mean by operands in microprocessor?
An opcode is a single instruction in assembly language. An operand is the data it does something with.For example, in "MOV r0, #0C", MOV is the opcode ("move this value into this register"), while r0 (register 0) and #0C (the number 12) are operands.
What is Op code and Operand?
An opcode (operation code) is a part of an instruction in machine language that specifies the operation to be performed, such as addition, subtraction, or data movement. The operand, on the other hand, is the part of the instruction that provides the necessary data or addresses required for the operation, indicating the target of the operation or the data to be manipulated. Together, the opcode and operand define the specific action and the data involved in machine-level programming.
How an assembler converts assembly mnemonics to machine language?
Each mnemonic maps directly to a machine instruction code, known as an opcode. Some mnemonics map to more than one opcode, however the instruction's operand types will determine which specific opcode will be generated.
What are operands?
Every instruction contains to parts: operation code[opcode],and operand. The first part of an instruction which specifies the task to be performed by the computer is called opcode. The second part of the instruction is the data to be operated on.,and it is called operand. The operand[or data]given in the instruction may be in various forms such as 8-bit or 16-bit data, 8-bit or 16-bit address, internal register or a register or memory location.
How do you use opcode and operand?
The hard way: Download the processor manuals and code the opcode and operands by hand The easy way: Use an assembler program. The instructions are slightly different for each program, so try reading the manuals.
What is addressing mode 8086?
Immediate addressing mode is when one of the operands is "immediately" located after the opcode. It is more correct to say that the operand is part of the instruction.
What is addressing mode in 8086?
Immediate addressing mode is when one of the operands is "immediately" located after the opcode. It is more correct to say that the operand is part of the instruction.
