(C) 尽管值比变量大小低一个数量级，但我似乎溢出了

Question

所以我正在为一个项目编写 C 入门类的代码(将小数转换为二进制(，并决定将二进制存储为十进制而不是使用数字数组。 该程序对于高达 1023 的数字运行良好，并且对于 1024 及以上的任何数字都让我一团糟

这是完整的代码(调用bclMain来运行它(：

void bclMain(void){
int d,in,in2;
do{
in=bclMenu();
switch(in){
case 1://dec to bin
do{
printf("tMENU CONVERSÃO DE DECIMAL PARA BINÁRIOnn");
printf("1 - 8 bits sem sinaln");
printf("2 - 8 bits sinal e módulon");
printf("3 - 8 bits complemento de 1n");
printf("4 - 8 bits complemento de 2n");
printf("5 - 16 bits sem sinaln");
printf("6 - 16 bits complemento de 2n");
printf("0 - Voltar ao menu anteriorn");
in2 = selectOp();
system("cls");
switch(in2){
case 1:
printf("8 bitstsem sinaln");
printf("Gama de valores [0,255]nn");
printf("valor a converter: ");
d=getNum(0,255);
printf("valor em binário: %dn",decToBin(d,0,8));
break;
case 2:
printf("8 bitstsinal e módulon");
printf("Gama de valores: [-127,127]nn");
printf("Valor a converter: ");
d=getNum(-127,127);
printf("Valor em binário: %d",decToBin(d,3,8));
break;
case 3:
printf("8 bitstcomplemento de 1n");
printf("Gama de valores: [-127,127]nn");
printf("Valor a converter: ");
d=getNum(-127,127);
printf("Valor em binário: %d",decToBin(d,2,8));
break;
case 4:
printf("8 bitstcomplemento de doisn");
printf("Gama de valores: [-128,127]nn");
printf("Valor a converter: ");
d=getNum(-128,127);
printf("Valor em binário: %d",decToBin(d,1,8));
break;
case 5:
printf("16 bitstsem sinaln");
printf("Gama de valores: [0,65535]nn");
printf("Valor a converter: ");
d=getNum(0,65535);
printf("Valor em binário: %llu",decToBin(d,0,16));
break;
case 6:
printf("16 bitstcomplemento de doisn");
printf("Gama de valores: [-32768,32767]nn");
printf("Valor a converter: ");
d=getNum(-32768,32767);
printf("Valor em binário: %llu",decToBin(d,1,16));
break;
}
if(in2!=0){
pause("nprima ENTER para continuarn");
system("cls");
}
}while(in2!=0);
break;
case 2:
do{
break;
}while(in2!=0);
case 0:
break;
default:
pause("Não implementado - prima ENTER para continuar");
break;
}
}while(in!=0);
}
int bclMenu(void){
int in;
system("cls");
printf("tCONVERSOR DE BASESn");
printf("1 - Decimalt-> Binárion");
printf("2 - Decimalt-> Hexadecimaln");
printf("3 - Bináriot-> Decimaln");
printf("4 - Bináriot-> Hexadecimaln");
printf("5 - Hexadecimalt-> Decimaln");
printf("6 - Hexadecimalt-> Binárion");
printf("0 - Voltar ao menu anteriorn");
in = selectOp();
system("cls");
return in;
}
unsigned long long int decToBin(int d,unsigned char mode,int digitNr){
//d is decimal value, mode indicates binary format to output in
unsigned long long int b=0;//b stands for binary, biggest int possible, should be able to store 19 digits (we needed at least 16)
unsigned char digits[16]={0},tf1; //tf1 stands for true/false 1
if(mode==0){//takes positive int and returns standard binary for it.
for(int cc=0;d>0;cc++){
//printf("%dn",(d%2) * intPow(10,cc));
b+=(d%2) * intPow(10,cc);
d= d/2;
}}
if(mode==1){//two's complement
if(d>=0)
b=decToBin(d,0,digitNr);
else{
b=decToBin(-d,0,digitNr);
for(int cc=0;b>0;cc++){
digits[cc]=b%10;
b=b/10;
}//standard binary separated into digits
//note - digits vector is backwards
tf1=0;
for(int cc=0;cc<digitNr;cc++){
//finding the first '1' and inverting everything after it
if(tf1){
if(digits[cc])
digits[cc]=0;
else
digits[cc]=1;
}
else if(digits[cc]==1)
tf1=1;
}
for(int cc=0;cc<digitNr;cc++){
b+= digits[cc] * intPow(10,cc);
//printf("digit - %d, multiplied - %dn",digits[cc],digits[cc] * intPow(10,cc));
}
}
}
if(mode==2){
if(d>=0)
b=decToBin(b,0,digitNr);
else{
b=decToBin(-b,0,digitNr);
for(int cc=0;b>0;cc++){
digits[cc]=b%10;
b=b/10;
}//standard binary separated into digits
//note - digits vector is backwards
for(int cc=0;cc<digitNr;cc++){
//invert all digits
if(digits[cc])
digits[cc]=0;
else
digits[cc]=1;
}
for(int cc=0;cc<digitNr;cc++){
//put digits back into a number
b+= digits[cc] * intPow(10,cc);
}
}
}
if(mode==3){
if(d>=0)
b=decToBin(d,0,digitNr);
else
b=decToBin(-d,0,digitNr)+intPow(10,digitNr);
}
return b;
}

这是实用工具库

void pause(char msg[]){
printf("%s",msg);
fflush(stdin);
getchar();
fflush(stdin);
}
unsigned long long int intPow(int a,unsigned int b){
int cc;
if(b==0)
return 1;
unsigned long long int c=a;
for (cc=1;cc<b;cc++)
a=a*c;//continuously multiply by base.
return a;
}
int selectOp(void){
int in;
fflush(stdin);
printf("Qual a sua opção?n");
scanf("%d",&in);
return in;
}
int getNum(int min,int max){
int in;
do{
scanf("%d",&in);
if((min<=in)&&(in<=max))
return in;
printf("valor inválidonnInsira novo valor: ");
}while((min>in)||(in>max));
}

如果这是太多的代码或不清楚的地方，我深表歉意，这是我的第一篇文章。

您需要的任何信息，只需询问

PS这是伦敦时间今天23：50到期的，我正在尽快通过这个

Answer 1

错误是 intintPow：

unsigned long long int c=a;
for (cc=1;cc<b;cc++)
a=a*c;//continuously multiply by base.
return a;

传入的参数a的类型为int，它是要返回的a的值。 因此，对于任何大于 10b，您都会溢出。 您希望c成为您累积并返回的变量：

for (cc=1;cc<b;cc++)
c=c*a;
return c;