[英]Hex/Dec Program in C getting the wrong output and cant use Scanf
每当我运行程序时,我都认为使用包含的测试字符串会得到错误的输出,尽管我认为我的第一个功能正在运行。 我拥有的文件是xbits.c xbits.h和showxbits.c的两个版本,一个是教师提供的版本,另一个是我试图与scanf一起使用的版本。 该程序应该将整数转换为十六进制字符串,然后将十六进制字符串转换为整数。 我的主要问题是,尽管我认为我的代码可用于讲师的测试输入,但我知道它不适用于scanf showxbits,因为当输入127时,它会给出诸如0xS的答案。
这是xbits.c
#include <stdio.h>
#include <math.h>
int hex_To_dec(int c) {
char hex_values[] = "aAbBcCdDeEfF";
int i;
int answer = 0;
for (i=0; answer == 0 && hex_values[i] != '\0'; i++) {
if (hex_values[i] == c) {
answer = 10 + (i/2);
}
}
return answer;
}
/* function represents the int n as a hexstring which it places in the
hexstring array */
void itox(char* s, int n)
{
char *digits = "0123456789ABCDEF";
int i=0,j;
char temp;
while(n > 0)
{
s[i] = digits[n % 16];
n /= 16;
i++;
}
s[i] = '\0'; // Add null terminator
i--;
// Now reverse it in place
for(j=0; j < i / 2; j++)
{
temp = s[j];
s[j] = s[i - j];
s[i - j] = temp;
}
}
/* function converts hexstring array to equivalent integer value */
int xtoi(char hexstring[]) {
//printf("in xtoi, processing %s\n", hexstring);
int answer = 0;
int i = 0;
int valid = 1;
int hexit;
if (hexstring[i] == '0') {
++i;
if (hexstring[i] == 'x' || hexstring[i] == 'X') {
++i;
}
}
while(valid && hexstring[i] != '\0') {
answer = answer * 16;
if(hexstring[i] >='0' && hexstring[i] <= '9') {
answer = answer + (hexstring[i] - '0');
}
else {
hexit = hex_To_dec(hexstring[i]);
if (hexit == 0) {
valid = 0;
}
else {
answer = answer + hexit;
}
}
++i;
}
if(!valid) {
answer = 0;
}
return answer;
}
这是讲师提供的showxbits.c:
/*
* stub driver for functions to study integer-hex conversions
*
*/
#include <stdio.h>
#include <string.h>
#include "xbits.h"
#define ENOUGH_SPACE 1000 /* not really enough space */
int main() {
char hexstring[ENOUGH_SPACE];
int m=0, n = 0x79FEB220;
itox(hexstring, n);
/* for stub testing: create a fake input string */
strcpy(hexstring, "6BCD7890");
m = xtoi(hexstring);
printf("\t%12d %s %12d\n", n, hexstring, m);
return 0; /* everything is just fine */
}
这是其中包含scanf的showxbits:
/*
* stub driver for functions to study integer-hex conversions
*
*/
#include <stdio.h>
#include <string.h>
#include "xbits.h"
#define ENOUGH_SPACE 100 /* not really enough space */
int main() {
char hexstring[ENOUGH_SPACE];
//int m=0, n = 0x79FEB220;
int n, m;
while ((scanf("%d", &n)) == 1) {
itox(hexstring, n);
m = xtoi( hexstring);
printf("%12d %s %12d\n", n, hexstring, m);
}
return 0; /* everything is just fine */
}
就像我说的那样,使用scanf函数时我得到了奇怪的输出。 我是一个完整的初学者程序员,非常感谢可以提供的任何帮助。 谢谢!
因为函数itox
存在错误,所以在反向字符串时会导致错误的结果。 然后,来自itox
的错误十六进制itox
将导致异常输出。
快速解决方案是将j < i / 2
替换为j < i / 2 + 1
void itox(char* s, int n)
{
//......
// Now reverse it in place
for(j=0; j < i / 2 + 1 ; j++)
{
temp = s[j];
s[j] = s[i - j];
s[i - j] = temp;
}
}
您无需反转字符串即可转换为十六进制ASCII:
#include <stdio.h>
const char* hexlat="0123456789ABCDEF";
char *binaryToHex(unsigned int answer, char *result){
if(answer==0) return result;
else{
result=binaryToHex(answer>>4,result);
*result=hexlat[answer & 0x0F];
return result+1;
}
};
int main(void) {
unsigned int answer=0x12340ADF;
char hexAnswer[32];
*binaryToHex(answer,hexAnswer)='\0';
printf("%s",hexAnswer);
return 0;
}
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.