私のマイクロプロセッサクラスのインストラクターが私たちに課題を与えて言った：
「Cでアセンブラーを作成します。」-私の最愛の教授
だから、私には少し非論理的に思えた。
私が間違っていなければ、アセンブリ言語は、機械語コードから高レベル言語の旅への最初のステップです。つまり、Cはアセンブリよりも高レベルの言語です。では、Cでアセンブラーを書く意味は何ですか？C言語がないのに、彼らは過去何をしていたのですか？彼らはマシンコードでアセンブラーを書いていましたか？
低レベル言語用の機械語翻訳を高レベル言語で書くのは意味がありません。
たとえば、そのアーキテクチャ用のCコンパイラさえない、まったく新しいマイクロプロセッサアーキテクチャを作成したとします。Cで記述されたアセンブラは、新しいアーキテクチャをシミュレートできますか？私はそれが役に立たないかどうかを意味しますか？
ところで、GNU AssemblerとNetwide AssemblerはCで書かれていることを知っています。また、なぜそれらがCで書かれているのだろうかと思っています。
最後に、これは私たちの教授が私たちに与えた単純なアセンブラのソースコードの例です。
// to compile, gcc assembler.c -o assembler
// No error check is provided.
// Variable names cannot start with 0-9.
// hexadecimals are twos complement.
// first address of the code section is zero, data section follows the code section.
//fout tables are formed: jump table, ldi table, label table and variable table.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
//Converts a hexadecimal string to integer.
int hex2int( char* hex)
{
int result=0;
while ((*hex)!='\0')
{
if (('0'<=(*hex))&&((*hex)<='9'))
result = result*16 + (*hex) -'0';
else if (('a'<=(*hex))&&((*hex)<='f'))
result = result*16 + (*hex) -'a'+10;
else if (('A'<=(*hex))&&((*hex)<='F'))
result = result*16 + (*hex) -'A'+10;
hex++;
}
return(result);
}
main()
{
FILE *fp;
char line[100];
char *token = NULL;
char *op1, *op2, *op3, *label;
char ch;
int chch;
int program[1000];
int counter=0; //holds the address of the machine code instruction
// A label is a symbol which mark a location in a program. In the example
// program above, the string "lpp", "loop" and "lp1" are labels.
struct label
{
int location;
char *label;
};
struct label labeltable[50]; //there can be 50 labels at most in our programs
int nooflabels = 0; //number of labels encountered during assembly.
// Jump instructions cannot be assembled readily because we may not know the value of
// the label when we encountered a jump instruction. This happens if the label used by
// that jump instruction appear below that jump instruction. This is the situation
// with the label "loop" in the example program above. Hence, the location of jump
// instructions must be stored.
struct jumpinstruction
{
int location;
char *label;
};
struct jumpinstruction jumptable[100]; //There can be at most 100 jumps
int noofjumps=0; //number of jumps encountered during assembly.
// The list of variables in .data section and their locations.
struct variable
{
int location;
char *name;
};
struct variable variabletable[50]; //There can be 50 varables at most.
int noofvariables = 0;
//Variables and labels are used by ldi instructions.
//The memory for the variables are traditionally allocated at the end of the code section.
//Hence their addresses are not known when we assemble a ldi instruction. Also, the value of
//a label may not be known when we encounter a ldi instruction which uses that label.
//Hence, the location of the ldi instructions must be kept, and these instructions must be
//modified when we discover the address of the label or variable that it uses.
struct ldiinstruction
{
int location;
char *name;
};
struct ldiinstruction lditable[100];
int noofldis=0;
fp = fopen("name_of_program","r");
if (fp != NULL)
{
while(fgets(line,sizeof line,fp)!= NULL) //skip till .code section
{
token=strtok(line,"\n\t\r ");
if (strcmp(token,".code")==0 )
break;
}
while(fgets(line,sizeof line,fp)!= NULL)
{
token=strtok(line,"\n\t\r "); //get the instruction mnemonic or label
//======================================== FIRST PASS ======================================================
while (token)
{
if (strcmp(token,"ldi")==0) //---------------LDI INSTRUCTION--------------------
{
op1 = strtok(NULL,"\n\t\r "); //get the 1st operand of ldi, which is the register that ldi loads
op2 = strtok(NULL,"\n\t\r "); //get the 2nd operand of ldi, which is the data that is to be loaded
program[counter]=0x1000+hex2int(op1); //generate the first 16-bit of the ldi instruction
counter++; //move to the second 16-bit of the ldi instruction
if ((op2[0]=='0')&&(op2[1]=='x')) //if the 2nd operand is twos complement hexadecimal
program[counter]=hex2int(op2+2)&0xffff; //convert it to integer and form the second 16-bit
else if (( (op2[0])=='-') || ((op2[0]>='0')&&(op2[0]<='9'))) //if the 2nd operand is decimal
program[counter]=atoi(op2)&0xffff; //convert it to integer and form the second 16-bit
else //if the second operand is not decimal or hexadecimal, it is a laber or a variable.
{ //in this case, the 2nd 16-bits of the ldi instruction cannot be generated.
lditable[noofldis].location = counter; //record the location of this 2nd 16-bit
op1=(char*)malloc(sizeof(op2)); //and the name of the label/variable that it must contain
strcpy(op1,op2); //in the lditable array.
lditable[noofldis].name = op1;
noofldis++;
}
counter++; //skip to the next memory location
}
else if (strcmp(token,"ld")==0) //------------LD INSTRUCTION---------------------
{
op1 = strtok(NULL,"\n\t\r "); //get the 1st operand of ld, which is the destination register
op2 = strtok(NULL,"\n\t\r "); //get the 2nd operand of ld, which is the source register
ch = (op1[0]-48)| ((op2[0]-48) << 3); //form bits 11-0 of machine code. 48 is ASCII value of '0'
program[counter]=0x2000+((ch)&0x00ff); //form the instruction and write it to memory
counter++; //skip to the next empty location in memory
}
else if (strcmp(token,"st")==0) //-------------ST INSTRUCTION--------------------
{
//to be added
}
else if (strcmp(token,"jz")==0) //------------- CONDITIONAL JUMP ------------------
{
//to be added
}
else if (strcmp(token,"jmp")==0) //-------------- JUMP -----------------------------
{
op1 = strtok(NULL,"\n\t\r "); //read the label
jumptable[noofjumps].location = counter; //write the jz instruction's location into the jumptable
op2=(char*)malloc(sizeof(op1)); //allocate space for the label
strcpy(op2,op1); //copy the label into the allocated space
jumptable[noofjumps].label=op2; //point to the label from the jumptable
noofjumps++; //skip to the next empty location in jumptable
program[counter]=0x5000; //write the incomplete instruction (just opcode) to memory
counter++; //skip to the next empty location in memory.
}
else if (strcmp(token,"add")==0) //----------------- ADD -------------------------------
{
op1 = strtok(NULL,"\n\t\r ");
op2 = strtok(NULL,"\n\t\r ");
op3 = strtok(NULL,"\n\t\r ");
chch = (op1[0]-48)| ((op2[0]-48)<<3)|((op3[0]-48)<<6);
program[counter]=0x7000+((chch)&0x00ff);
counter++;
}
else if (strcmp(token,"sub")==0)
{
//to be added
}
else if (strcmp(token,"and")==0)
{
//to be added
}
else if (strcmp(token,"or")==0)
{
//to be added
}
else if (strcmp(token,"xor")==0)
{
//to be added
}
else if (strcmp(token,"not")==0)
{
op1 = strtok(NULL,"\n\t\r ");
op2 = strtok(NULL,"\n\t\r ");
ch = (op1[0]-48)| ((op2[0]-48)<<3);
program[counter]=0x7500+((ch)&0x00ff);
counter++;
}
else if (strcmp(token,"mov")==0)
{
//to be added
}
else if (strcmp(token,"inc")==0)
{
op1 = strtok(NULL,"\n\t\r ");
ch = (op1[0]-48)| ((op1[0]-48)<<3);
program[counter]=0x7700+((ch)&0x00ff);
counter++;
}
else if (strcmp(token,"dec")==0)
{
//to be added
}
else //------WHAT IS ENCOUNTERED IS NOT AN INSTRUCTION BUT A LABEL. UPDATE THE LABEL TABLE--------
{
labeltable[nooflabels].location = counter; //buraya bir counter koy. error check
op1=(char*)malloc(sizeof(token));
strcpy(op1,token);
labeltable[nooflabels].label=op1;
nooflabels++;
}
token = strtok(NULL,",\n\t\r ");
}
}
//================================= SECOND PASS ==============================
//supply the address fields of the jump and jz instructions from the
int i,j;
for (i=0; i<noofjumps;i++) //for all jump/jz instructions
{
j=0;
while ( strcmp(jumptable[i].label , labeltable[j].label) != 0 ) //if the label for this jump/jz does not match with the
j++; // jth label in the labeltable, check the next label..
program[jumptable[i].location] +=(labeltable[j].location-jumptable[i].location-1)&0x0fff; //copy the jump address into memory.
}
// search for the start of the .data segment
rewind(fp);
while(fgets(line,sizeof line,fp)!= NULL) //skip till .data, if no .data, also ok.
{
token=strtok(line,"\n\t\r ");
if (strcmp(token,".data")==0 )
break;
}
// process the .data segment and generate the variabletable[] array.
int dataarea=0;
while(fgets(line,sizeof line,fp)!= NULL)
{
token=strtok(line,"\n\t\r ");
if (strcmp(token,".code")==0 ) //go till the .code segment
break;
else if (token[strlen(token)-1]==':')
{
token[strlen(token)-1]='\0'; //will not cause memory leak, as we do not do malloc
variabletable[noofvariables].location=counter+dataarea;
op1=(char*)malloc(sizeof(token));
strcpy(op1,token);
variabletable[noofvariables].name=op1;
token = strtok(NULL,",\n\t\r ");
if (token==NULL)
program[counter+dataarea]=0;
else if (strcmp(token, ".space")==0)
{
token=strtok(NULL,"\n\t\r ");
dataarea+=atoi(token);
}
else if((token[0]=='0')&&(token[1]=='x'))
program[counter+dataarea]=hex2int(token+2)&0xffff;
else if (( (token[0])=='-') || ('0'<=(token[0])&&(token[0]<='9')) )
program[counter+dataarea]=atoi(token)&0xffff;
noofvariables++;
dataarea++;
}
}
// supply the address fields for the ldi instructions from the variable table
for( i=0; i<noofldis;i++)
{
j=0;
while ((j<noofvariables)&&( strcmp( lditable[i].name , variabletable[j].name)!=0 ))
j++;
if (j<noofvariables)
program[lditable[i].location] = variabletable[j].location;
}
// supply the address fields for the ldi instructions from the label table
for( i=0; i<noofldis;i++)
{
j=0;
while ((j<nooflabels)&&( strcmp( lditable[i].name , labeltable[j].label)!=0 ))
j++;
if (j<nooflabels){
program[lditable[i].location] = (labeltable[j].location)&0x0fff;
printf("%d %d %d\n", i, j, (labeltable[j].location));
}
}
//display the resulting tables
printf("LABEL TABLE\n");
for (i=0;i<nooflabels;i++)
printf("%d %s\n", labeltable[i].location, labeltable[i].label);
printf("\n");
printf("JUMP TABLE\n");
for (i=0;i<noofjumps;i++)
printf("%d %s\n", jumptable[i].location, jumptable[i].label);
printf("\n");
printf("VARIABLE TABLE\n");
for (i=0;i<noofvariables;i++)
printf("%d %s\n", variabletable[i].location, variabletable[i].name);
printf("\n");
printf("LDI INSTRUCTIONS\n");
for (i=0;i<noofldis;i++)
printf("%d %s\n", lditable[i].location, lditable[i].name);
printf("\n");
fclose(fp);
fp = fopen("RAM","w");
fprintf(fp,"v2.0 raw\n");
for (i=0;i<counter+dataarea;i++)
fprintf(fp,"%04x\n",program[i]);
}
}
2
分離されたデバイスは存在しません。クロスツールチェーンは、特に小さなアーキテクチャで非常に普及しています。
—
ラースヴィクランド16
「クロス」コンパイラ/アセンブラは、ターゲットとは異なるシステムで実行され、ターゲットシステムでの使用に適したアーティファクトを生成します。古代では、システム間のデータ交換は必ずしも必要ではありませんでしたが、システム自体をゼロからブートストラップする必要がありました。アーキテクチャのほとんどすべての最新の開発は、確立されたシステムで行われ、すべてをクロスコンパイルします。
—
ラースヴィクランド16
Cではなくマシンコードでアセンブラを記述しますか？あなたの教授はあなたに親切です。
—
ウィンストンユワート
可能な限り最高のプログラミング環境/言語ですべてのコードを記述しようとしないのはなぜですか？アセンブラも例外ではありません。
—
エリックエイド16
特定の方向に決まった「旅」はありません。
—
-whatsisname