代码中有一些人类可读代码的注释：
#include <stdio.h>
#include <string.h>
#define SIZE 100 //size of the input array and output array
#define ACCUM_CHAR_SIZE 25 //size of the temp array
int main(){
char i[SIZE];
char acc[ACCUM_CHAR_SIZE];
char o[SIZE];
int it_l = 0, it_a = 0, it_r = 0;
//it_l is the iterator to the input sentence,
//it_a is the iterator to the temp array
//it_r is the iterator to the output sentence
printf("Enter a sentence:");
gets(i);
int len = strlen(i) - 1;
while(it_l <= len){
if(i[len - it_l] != ' '){
acc[it_a] = i[len - it_l]; //add letters to acc until space
it_a++;
}
else{
it_a -= 1;
//acc is reversed, I reversed it again to the output sentence
while(it_a >= 0){
o[it_r] = acc[it_a];
it_r++;
it_a--;
}
it_r += 1;
o[it_r] = 32; //put a space
it_a = 0; //reset the temp array
strcpy(acc, ""); //clear the acc
}
it_l++;
}
printf("%s", o);
}
程序理论上看起来不错，但是当它被执行时，它有时会打印垃圾值，只有一些单词，或者句子，这些句子只用垃圾值而不是空格反转了一半。
上面的程序是将每个单词保存到 temp，并将反向 temp(存储单词时 temp 反转(保存回输出。但是，它失败了。
谢谢你的帮助。
#include <stdio.h>
#include <string.h>
#define SIZE 100 //The size of input array is prefered to be equal to ouput array.
int main(){
char input[SIZE];
char output[SIZE];
int i = 0, j = 0;
//i is the iterator to the input sentence,
//j is the iterator to the output sentence
printf("Enter a sentence:");
gets(input);
int len = strlen(input) - 1; //Total length.
j = len;
while(input[i]!= NULL){
output[j] = input[i];
i++;
j--;
}
output[len+1]= NULL;
printf("%s", output);
printf("Finished");
}
至少有三个问题。
第一个问题是你永远不会终止字符串o要执行该更改：
printf("%s", o);
到
o[it_r] = ' ';
printf("%s", o);
第二个问题是您不正确地递增it_r。改变
it_r += 1;
o[it_r] = 32; //put a space
到
o[it_r] = ' '; // Use ' ' instead of 32
it_r += 1;
第三个问题是你不处理输入的第一个单词(因为前面没有空格(。我会把这个问题留给你作为练习。
顺便说一句：不要使用gets来读取输入。请改用fgets。
尝试下面给出的修改后的代码，更改的部分已被注释(为了可读性，所有其他注释都被删除了(
#include <stdio.h>
#include <string.h>
#define SIZE 100
#define ACCUM_CHAR_SIZE 25
int main(){
char i[SIZE];
char acc[ACCUM_CHAR_SIZE];
char o[SIZE];
int it_l = 0, it_a = 0, it_r = 0;
printf("Enter a sentence:");
gets(i);
int len = strlen(i) - 1;
while(it_l <= len){
if(i[len - it_l] != ' '){
acc[it_a] = i[len - it_l];
it_a++;
}
else{
it_a -= 1;
while(it_a >= 0){
o[it_r] = acc[it_a];
it_r++;
it_a--;
}
/*it_r += 1; There is no need to increment it_r here as
it is already incremented in the above loop*/
o[it_r] = 32;
it_r += 1; /* The above increment statement was moved here(only
incremented when a new value is loaded in to output array) */
it_a = 0;
strcpy(acc, "");
}
it_l++;
}
/* The below section reverses and stores the first word of the
input which was not getting operated in the above loop */
it_a -= 1;
while(it_a >= 0){
o[it_r] = acc[it_a];
it_r++;
it_a--;
}
o[it_r] = ' '; // Terminating output array
printf("%s", o);
}
上面的代码将按预期工作，但有一些小问题(如下所述(
不建议使用 gets((：- get 输入字符串。有关详细信息，请参阅此链接。
临时数组的大小acc：- 如果输入大于 25 个字符的单词，程序可能会输出垃圾值。
没有临时存储的版本。
- 从"开始"屏幕读取输入字符串 (0(
- 从输出的末尾开始，将输入中的任何单词复制到输出中 - 单词不会反转
- 从输出中的位置复制输出中的任何空格，从末尾开始
法典
#define SIZE 100 //size of the input array and output array
int main(){
char i[SIZE];
char o[SIZE];
printf("Enter a sentence: ");
fgets(i, SIZE, stdin); // fgets instead of gets
int j,len = strlen(i) - 1; // assuming you eat the EOL character
o[len] = ' '; // start by marking end of output string
for(j=0 ; j<len ; j++) { // navigating the input string
if (i[j] == ' ') {
o[len-j-1] = i[j]; // if space, copy (from end) in output
}
else {
int k=j;
do {
k++; // count(+1) the word, non-space, characters
} while (k<len && i[k] != ' ');
strncpy(&o[len-k], &i[j], k-j); // copy the word
j = k-1;
}
}
printf("%sn", o);
}