[英]Counting suitable first letters of the words in string in C language
请帮助我的部分功能。 在函数中,我创建了一个for循环,该循环应该计算字符串中每个单词开头的合适字母。 例如,如果我将字符串“ Beautiful skyes”传递给函数和char's',则for循环后的变量“ count”必须等于1,但这不会发生。 有什么问题? 我在编码方面是个新手,但是这对我来说是正确的,但是由于某种原因,这个for循环无法按我的预期工作(这只是整个函数的一部分,但是问题出在for循环,因为即使有一些合适的字母,它也总是返回NULL):
char** rearrange_string(char *str, char letter, int *size)
{
char *search, **array, upper_letter = toupper(letter), *shift;
int count = 0, i=0, j=0, counter;
for (search = str; *search != '\0'; search++) {
if (*search == ' ') {
continue;
}
if (*search != letter || *search != upper_letter) {
while (*search != ' ') {
search++;
if (*search == '\0') {
break;
}
}
continue;
}
if (*search == letter || *search == upper_letter) {
count++;
while (*search != ' ') {
search++;
if (*search == '\0') {
break;
}
}
}
}
if (count == 0) {
printf("There is no suitable data. Please reinput the string.");
return NULL;
}
*search != letter || *search != upper_letter
如果letter != upper_letter
始终为true
更改为:
*search != letter && *search != upper_letter
当使用字符串"Beautiful skies"
运行时,该程序会出现缓冲区溢出,因为search
不在原始字符串之内。 在这种情况下*search == letter
你增量search
,直到*search==0
,然后search
在循环结束时再次增加,所以它超越了分配的缓冲区。
以下分析显示:
多谢你们! 问题出在for循环的第一个if语句中。 我应该用&&代替||。 谢谢! 如果有人想看整个程序,这里是一个完整的代码:
#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#include <ctype.h>
char** rearrange_string(char *str, char letter, int *size);
void main()
{
int size=0, i=0;
char *str, letter, **array;
printf("Enter the string and press ENTER to check: ");
gets(str);
printf("Enter the character for check (only lower chars!): ");
scanf("%c", &letter);
array = rearrange_string(str, letter, &size);
if (array == NULL)
{
printf("\nNo data found, please relaunch the program.");
return;
}
printf("The output string(s) is(are):\n");
for (i=0; i<size; i++)
{
printf("%s", array[i]);
printf("\n");
}
printf("size: %d", size);
for (i=0; i<size; i++)
{
free(array[i]);
}
free(array);
}
char** rearrange_string(char *str, char letter, int *size)
{
char *search, **array, upper_letter = toupper(letter), *shift;
int count = 0, i=0, j=0, counter;
for (search = str; *search != '\0'; search++)
{
if (*search == ' ')
{
continue;
}
if (*search != letter && *search != upper_letter)
{
while (*search != ' ')
{
search++;
if (*search == '\0')
{
break;
}
}
continue;
}
if (*search == letter || *search == upper_letter)
{
count++;
while (*search != ' ')
{
search++;
if (*search == '\0')
{
break;
}
}
}
}
if (count == 0)
{
printf("There is no suitable data. Please reinput the string.");
return NULL;
}
*size = count;
array = (char**)malloc(count*sizeof(char*));
for (search = str; *search != '\0'; search++)
{
if (*search == ' ')
{
continue;
}
if (*search != letter && *search != upper_letter)
{
while (*search != ' ')
{
search++;
if (*search == '\0')
{
break;
}
}
continue;
}
if (*search == letter || *search == upper_letter)
{
counter = 1;
j=0;
shift = search;
while (*search != ' ')
{
counter++;
search++;
if (*search == '\0')
{
break;
}
}
array[i] = (char*)malloc(counter*sizeof(char));
while (*shift != ' ')
{
*(*(array+i)+j) = *shift;
j++;
shift++;
if (*shift == '\0')
{
break;
}
}
*(*(array+i)+j) = '\0';
i++;
}
}
return array;
}
像这样的函数呢?
#include <unistd.h>
#include <stdio.h>
size_t count_char(char *str, char s)
{
size_t count = 0;
for (size_t i = 0; str[i]; i++)
if (str[i] == s)
count += 1;
return (count);
}
此代码将仅返回在字符串中找到的字符数,例如:
printf(“%li \\ n”,count_char(“美丽的天空”,'s'));
将打印:
1
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