我不知道为什么我的功能没有给出正确的结果。我怀疑它没有返回正确的类型(unsigned long long int),但它返回一个int。
#include <stdlib.h>
#include <stdio.h>
#include <math.h>
// compile with:
// gcc prog_long_long.c -o prog_long_long.exe
// run with:
// prog_long_long
unsigned long long int dec2bin(unsigned long long int);
int main() {
unsigned long long int d, result;
printf("Enter an Integer\n");
scanf("%llu", &d);
result = dec2bin(d);
printf("The number in binary is %llu\n", result);
system("pause");
return 0;
}
unsigned long long int dec2bin(unsigned long long int n) {
unsigned long long int rem;
unsigned long long int bin = 0;
int i = 1;
while (n != 0) {
rem = n % 2;
n = n / 2;
bin = bin + (rem * i);
i = i * 10;
}
return bin;
}
以下是具有不同输入值的结果的输出:
C:\Users\desktop\Desktop\gcc prog_long_long.c -o prog_long_long.exe
C:\Users\desktop\Desktop\prog_long_long
Enter an Integer
1023
The number in binary is 1111111111
The number in octal is 1777
C:\Users\desktop\Desktop\prog_long_long
Enter an Integer
1024
The number in binary is 1410065408
The number in octal is 2994
答案 0 :(得分:1)
您不能以这种方式将数字转换为二进制,十进制和二进制是相同数字的外部表示。您应该将数字转换为C字符串,从右到左一次计算一个二进制数字。
以下是64位长long的工作原理:
#include <stdio.h>
#include <string.h>
char *dec2bin(char *dest, unsigned long long int n);
int main(void) {
unsigned long long int d;
char buf[65], *result;
printf("Enter an Integer\n");
if (scanf("%llu", &d) == 1) {
result = dec2bin(buf, d);
printf("The number in binary is %s\n", result);
}
//system("pause");
return 0;
}
char *dec2bin(char *dest, unsigned long long int n) {
char buf[65];
char *p = buf + sizeof(buf);
*--p = '\0';
while (n > 1) {
*--p = (char)('0' + (n % 2));
n = n / 2;
}
*--p = (char)('0' + n);
return strcpy(dest, p);
}