问题
I have just recently started learning C. I wrote a very short program that converts between decimal and binary. I wanted to try and write a code that converts between decimal and any base (up until 36). However, my code just prints out garbage.
#include <stdio.h>
#include <string.h>
void printBase(int n, int k, int i, char a[])
{
int placeholder;
if (n != 0)
{
//return n % 2 + 10 * printBinary(n / 2);
placeholder=(char)(n%k);
if(placeholder>=10)
{
a[i] = (char)(placeholder - 10) + 'A';
} else {
a[i] = (char)placeholder;
}
i++;
printBase(n/2, k, i, a);
}
for (i=0; a[i]!='\0'; i++)
{
printf("%c", a[i]);
}
return;
}
void reverse(char fromStr[], char toStr[])
{
int i, j=0;
i=getchar();
for (i=0; fromStr[i]!='\0'; i++)
{
j++;
}
i=0;
while (j>=0)
{
toStr[i]=fromStr[j];
i++;
j--;
}
printf("%s", toStr);
}
int main()
{
int n, k;
char a[81], b[81];
setvbuf(stdout, NULL, _IONBF, 0);
printf("Enter a deicmal number you want to convert to binary: ");
scanf("%i", &n);
fflush(stdout);
printf("Enter a base: ");
scanf("%i", &k);
printBase(n, k, 0, a);
//printf("%s", a);
//reverse(a, b);
return 0;
}
I thought the problem was with my reverse function but it works fine outside of this code. Even when I print out string a inside the printBase function it prints out garbage. What is the problem here?
回答1:
Based on your code, the following does what you want. It places in a
a reverse conversion that must still be printed backwards:
void convertBase(int n, int k, char a[])
{
int j, i=0, sign= 0;
if (n==0) a[i++]='0';
if (n<0 ) {sign= -1; n= -n;}
while (n>0) {
j= n%k;
if (j<10)
a[i]= j+'0';
else
a[i]= j+'A';
n= n/k;
i++;
}
if (sign== -1) a[i++]= '-';
a[i]= 0;
}
And here is revert:
void revStr(char *s)
{
char c;
int i=0, j=0;
while (s[i]) i++; i--;
while (i>j) {
c= s[j];
s[j]=s[i];
s[i]= c;
i--; j++;
}
}
回答2:
With some general implementation of itoa
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <limits.h>
static const char* DIGISTS = "0123456789ABCDEFGHIKLMNOPQRSTVXYZ";
void itoa(long long i, unsigned char radix, char* to) {
char *s = to + 65;
char sign;
unsigned len;
if(i < 0) {
sign = 1;
len = 2;
} else {
sign = 0;
len = 1;
}
*s = '\0';
do {
*(--s)= * ( DIGISTS + abs(i % radix) );
i /= radix;
++len;
} while(i != 0);
if(sign)
*(--s) = '-';
memmove( to, s, len );
}
int main(int argc, const char** argv)
{
char a[65];
itoa( LLONG_MAX, 2, a);
printf("binary: %s \n", a);
itoa(12345, 10, a);
printf("digit: %s \n", a);
itoa(64018, 16, a);
printf("hex : 0x%s \n", a);
itoa(-24, 24, a);
printf("base 24 : base24x%s \n", a);
itoa(LLONG_MAX, 36, a);
printf("base 36 : base36x%s \n", a);
return 0;
}
回答3:
#define c ('A' - 10)
int n = 4564567; // any number
int b = 16; // any base
int tmp = n < 0 ? -n : n;
int i = 0;
while (tmp)
{
tmp /= b;
++i; // you need to calculate how long will be the number
}
while (i--)
{
a[i] = n % b + ((n % b < 10) ? '0' : c); // you have to check if the remaining is below 10 or not. That is very important
n /= b;
}
I think you forgot to add '0' in the else case
Adapted from an old source file https://github.com/agavrel/42-ft_printf/blob/master/srcs/pf_number.c
回答4:
An alternative to reversing the array uses recursion.
As the number of digits in some integer type in never so great, recursion will not be excessive.
The below used a do {} while
loop to avoid a special case with printBase(0)
.
It also uses negative values in printBase_helper()
where /,%
is well defined with C99. This avoids undefined behavior of code like n = -n
when n == INT_MIN
.
void printBase_helper(int neg, int base) {
if (neg <= -base) {
printBase_helper(neg / base, base);
neg %= base;
}
putchar("0123456789ABCDEFGHIJKLMONOPQSTUVWXYZ"[-neg]);
}
void printBase(int n, int base) {
if (n < 0) {
putchar('-');
} else {
n = -n;
}
printBase_helper(n, base);
}
Test code
int main(void) {
int value[] = {0, 1, -1, 10, INT_MAX, INT_MIN};
int base[] = {10, 2, 36};
for (unsigned v = 0; v < sizeof value / sizeof value[0]; v++) {
for (unsigned b = 0; b < sizeof base / sizeof base[0]; b++) {
printf("Base %2d, value %11d, --> base %2d, value ", 10, value[v], base[b]);
printBase(value[v], base[b]);
printf("\n");
}
}
return 0;
}
Output
Base 10, value 0, --> base 10, value 0
Base 10, value 0, --> base 2, value 0
Base 10, value 0, --> base 36, value 0
Base 10, value 1, --> base 10, value 1
Base 10, value 1, --> base 2, value 1
Base 10, value 1, --> base 36, value 1
Base 10, value -1, --> base 10, value -1
Base 10, value -1, --> base 2, value -1
Base 10, value -1, --> base 36, value -1
Base 10, value 10, --> base 10, value 10
Base 10, value 10, --> base 2, value 1010
Base 10, value 10, --> base 36, value A
Base 10, value 2147483647, --> base 10, value 2147483647
Base 10, value 2147483647, --> base 2, value 1111111111111111111111111111111
Base 10, value 2147483647, --> base 36, value ZIK0ZJ
Base 10, value -2147483648, --> base 10, value -2147483648
Base 10, value -2147483648, --> base 2, value -10000000000000000000000000000000
Base 10, value -2147483648, --> base 36, value -ZIK0ZK
回答5:
I didn't know that answers to my own question should be posted below, sorry! But here is the code that I wrote on my own if it helps anyone:
#include <stdio.h>
#include <string.h>
void printBinary(int n, int k, int i, char a[])
{
int placeholder;
if (n != 0)
{
placeholder=(n%k);
if(placeholder>=10)
{
a[i] = (placeholder - 10) + 'A';
} else if(placeholder>=0 && placeholder<=9){
a[i] = placeholder + '0';
}
i++;
a[i]='\0';
printBinary(n/k, k, i, a);
}
return;
}
int main()
{
int n, k;
char a[81], b[81];
setvbuf(stdout, NULL, _IONBF, 0);
printf("Enter a deicmal number you want to convert to binary: ");
scanf("%i", &n);
fflush(stdout);
printf("Enter a base: ");
scanf("%i", &k);
printBinary(n, k, 0, a);
n=strlen(a);
k=0;
n--;
while (n>=0)
{
b[k]=a[n];
n--;
k++;
}
b[k]='\0';
printf("%s", b);
return 0;
}
回答6:
Try this one.
#include<stdio.h>
int main()
{
int n,b,t1,i=0;
float t2,t3,t4;
int a[10];
scanf("%d%d",&n,&b);
for( i=0; n!=0; i++)
{
t1=n/b;
t2=(float)n/b;
t3=(float)t2-t1;
t1=t3*b;
t2=(float)t3*b;
t4=(float)t2-t1;
if(t4>=0.5)
{
a[i]=t3*b+1;
}
else
{
a[i]=t3*b;
}
n=n/b;
}
for(int j=i; j>0; j--)
{
printf("%d",a[j-1]);
}
return 0;
}
来源:https://stackoverflow.com/questions/52932715/c-code-for-converting-decimal-to-any-base-from-2-to-36