我知道这个问题之前已经被问过很多次了,但我就是无法理解我做错了什么。每当我取得一些进展时,我就会收到一个新错误。我使用的代码非常基本,因为我是新手,我们的教授要求使用 scanf 和 gets。这是我到目前为止的代码:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX_SIZE 100
int identify(char[], char[]);
int remove(char[], char[], int);
int scan(choice)
{
while(choice < 0 || choice > 7)
{
printf("Invalid input, choose again\n");
scanf("%d", &choice);
}
return choice;
}
int main()
{
char sentence[MAX_SIZE], word[MAX_SIZE];
int choice, i, j, k, deikths;
printf("Choose one of the following:\n");
printf("1. Give sentence\n");
printf("2. Subtract a word\n");
printf("3. Add a word\n");
printf("4. Count the words\n");
printf("5. Count the sentences\n");
printf("6. Count the characters\n");
printf("7. Is the phrase a palindrome?\n");
printf("0. Exit\n");
scanf("%d", &choice);
if(scan(choice) == 1)
{
printf("Give sentence:\n");
gets(sentence);
gets(sentence);
printf("%s\n", sentence);
}
else(scan(choice) == 2);
{
printf("Give word you want to subtract\n");
gets(word);
printf("%s", word);
deikths = identify(sentence, word);
if(deikths != -1)
{
remove(sentence, word, deikths);
printf("Sentence without word: %s\n", sentence);
}
else
{
printf("Word not found in sentence.\n");
}
}
}
int identify(char sentence[], char word[])
{
int i, j, k;
for(k = 0; word[k] != '\0'; k++);
{
for(i = 0, j = 0; sentence[i] != '\0'; i++)
{
if(sentence[i] == word[j])
{
j++;
}
else
{
j = 0;
}
}
}
if(j == 1)
{
return(i - j);
}
else
{
return -1;
}
}
int remove(char sentence[], char word[], int deikths)
{
int i, k;
for(k = 0; word[k] != '\0'; k++)
{
for(i = deikths; sentence[i] != '\0'; i++)
{
sentence[i] = sentence[i + k + 1];
}
}
}
我收到的错误是删除函数具有冲突的类型。任何有关修复我的代码的帮助将不胜感激,甚至对我的问题的替代解决方案也会很棒。
最佳答案
正如注释中所确定的,由于 remove
已在 stdio.h
中定义,因此会生成编译器错误。更改后,代码编译成功,但仍然无法按预期工作。
identify
是一个函数,用于查找字符串中是否存在子字符串并返回其位置。这与标准库中的 strstr
的工作方式非常相似 - 我建议查看该函数的实现,以更好地理解这是如何完成的。
您实现的函数只能正确查找字符串末尾长度为 1 的子字符串。我在下面的代码中突出显示了导致此问题的错误。
int identify(char sentence[], char word[])
{
int i, j, k;
for(k = 0; word[k] != '\0'; k++); // <- this loops is never actually ran because of the trailing semicolon - this is however a good thing as it is redundant
{
for(i = 0, j = 0; sentence[i] != '\0'; i++)
{
if(sentence[i] == word[j])
{
j++;
}
else
{
j = 0; // <- this makes it so only matches at the end can be found - otherwise, j is just reset back to 0
}
}
}
if(j == 1) // <- this makes it so only matches of length 1 can be found
{
return(i - j); // <- this is only correct if the match is at the end of the sentence
}
else
{
return -1;
}
}
strremove
由于嵌套循环而效率低下,并且需要缩短复制的字符范围 - 现在数据访问超出了数组末尾。
int strremove(char sentence[], char word[], int deikths)
{
int i, k;
for(k = 0; word[k] != '\0'; k++) // <- this loop is redundant
{
for(i = deikths; sentence[i] != '\0'; i++) // <- you need to add range checking to make sure sentence[i+k+1] doesn't go beyond the end of the string
{
sentence[i] = sentence[i + k + 1];
}
}
}
我会将 main
中的问题作为练习留给您 - 毕竟这是一项作业。
关于c - 尝试从 C 中的字符串中删除子字符串,但始终失败,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/53381873/