[英]Swap occurrences of two most frequent letters in a string
我不知道代码中的问题是什么,但是当我编译时会得到:
warning: passing arg 2 of `strcspn' makes pointer from integer without a cast
这是代码:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define STR_LEN 50
int main(void) {
int i = 0, j = 0, length = 0, count1 = 0, count2 = 0, count3 = 0;
char letter3 = 'a', letter2 = 'a', string[STR_LEN] = { 0 };
length = strlen(string);
printf("Enter a sentence: ");
fgets(string, STR_LEN, stdin);
for (i = 0; i < length; i++) {
for (j = 0; j < length; j++) {
if (string[i] == string[j]) {
count1++;
} else {
count1 = 0;
}
}
if (count1 > count3) {
count2 = count3;
count3 = count1;
letter2 = letter3;
letter3 = string[i];
} else
if (count1 > count2) {
count2 = count1;
letter2 = string[i];
}
}
string[strcspn(string, letter2)] = letter3;
string[strcspn(string, letter3)] = letter2;
printf("\n %s", string);
system("pause");
return 0;
}
该代码应该从用户那里获得一个句子,并用第二个公共字母切换该句子中最常见的字母。
strcspn()
函数采用两个字符串作为参数,但是您要传递一个字符串和一个字符。 您需要以某种方式将字符转换为字符串。 一种方法是:
int sep[2] = "";
sep[0] = letter2;
string[strcspn(string, sep)] = letter3;
sep[0] = letter3;
string[strcspn(string, sep)] = letter2;
但是,第一个调用将第一次出现的letter2
为letter3
; 第二个呼叫改变的第一次出现letter3
(这可能是刚刚更换的先前调用的)与letter2
。 这不是完成字符串转换的完整工作,您需要扫描整个字符串以进行更改。
一种可能性是这样的:
#include <ctype.h>
#include <stdio.h>
#include <string.h>
#define NULL_VALUE '\0'
static inline void map(char *str, int len, int c_old, int c_new)
{
for (int i = 0; i < len; i++)
{
if (str[i] == c_old)
str[i] = c_new;
}
}
int main(void)
{
char buffer[4096];
printf("Enter a sentence: ");
if (fgets(buffer, sizeof(buffer), stdin) == 0)
return 0;
int length = strlen(buffer);
if (length > 0)
buffer[--length] = '\0';
putchar('\n');
printf("Original [%s]\n", buffer);
int count[256] = { 0 };
for (int i = 0; i < length; i++)
{
if (isalpha((unsigned char)buffer[i]))
count[(unsigned char)buffer[i]]++;
}
int max1_count = 0;
int max2_count = 0;
char max1_value = '\0';
char max2_value = '\0';
for (int i = 0; i < 256; i++)
{
if (count[i] > max1_count)
{
max2_count = max1_count;
max2_value = max1_value;
max1_count = count[i];
max1_value = i;
}
else if (count[i] > max2_count)
{
max2_count = count[i];
max2_value = i;
}
}
/*
** Since a string is a sequence of non-null character codes followed
** by a null byte, it is safe to use '\0' as the temporary value in
** the three-step swap operation
*/
if (max2_count > 0)
{
map(buffer, length, max1_value, NULL_VALUE);
map(buffer, length, max2_value, max1_value);
map(buffer, length, NULL_VALUE, max2_value);
}
printf("Revised [%s]\n", buffer);
return 0;
}
使用宏NULL_VALUE
的唯一原因是三条map()
线的对称性不言而喻。
我调用了程序ccswap19
,并使用Bash的“这里字符串”来提供数据— putchar('\\n');
意味着输出出现在与提示分开的一行上。 如果您以交互方式运行程序,则在“原始”打印之前会有一个空白行。
$ ccswap19 <<< "The hidden costs of the exodus are now revealed for all to see."
Enter a sentence:
Original [The hidden costs of the exodus are now revealed for all to see.]
Revised [Tho hiddon cests ef tho oxedus aro new rovoalod fer all te soo.]
$ ccswap19 <<< "aaaaaaaaaaaa"
Enter a sentence:
Original [aaaaaaaaaaaa]
Revised [aaaaaaaaaaaa]
$ ccswap19 <<< "aaaabaaaaaaa"
Enter a sentence:
Original [aaaabaaaaaaa]
Revised [bbbbabbbbbbb]
$
strcspn(const char * str1,const char * str2)在第一个字符串的第二个字符串中查找任何字符的第一个实例。 您将传递字符而不是字符串作为第二个参数。 您需要函数strchr(const char * string,int character),该函数查找单个字符。
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