为什么此代码产生无限循环？

发布于 2025-01-22 09:43:36 字数 491 浏览 0 评论 0原文

#include <Stdio.h>
#include <string.h>

int main(){
    char str[51];
    int k = 1;
    printf("Enter string\n");
    scanf("%s", &str);
    for(int i = 0; i < strlen(str); i++){
        while(str[k] != '\0')){
            if(str[i] == str[k]){
                printf("%c", str[i]);
                k++;
            }
        }
    }

    return 0;
}

这是简单的C代码，检查字符串中的重复字符并打印字符。我不明白为什么它会产生无限的循环。内部循环应在str [k]到达零终端时停止，但程序无限继续。

原文

#include <Stdio.h>
#include <string.h>

int main(){
    char str[51];
    int k = 1;
    printf("Enter string\n");
    scanf("%s", &str);
    for(int i = 0; i < strlen(str); i++){
        while(str[k] != '\0')){
            if(str[i] == str[k]){
                printf("%c", str[i]);
                k++;
            }
        }
    }

    return 0;
}

It is simple C code that checks for duplicate characters in string and prints the characters. I am not understanding why it is producing an infinite loop. The inner while loop should stop when str[k] reaches the null terminator but the program continues infinitely.

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夏の忆 2025-01-29 09:43:36

要知道

您不需要将变量str的地址传达到scanf（）
不使用“％s”，使用“％＆lt; w width＆gt; s”，避免 buffer-overfloflfloflfloflfloflflofl
代码> scanf（）转换是否成功，通过检查其返回值
始终使用size_t通过任何数组
i＆lt; strlen（str），使循环的时间复杂性 o（n ³），而不是 o（n ²），这也不是很好，您应该检查str [i]！= 0。但是，c的许多现代编译器都会顺其自然。
#include＆lt; stdio.h＆gt;这是非常错误的，stdio.h！= stdio.h
call to printf（）可以使用puts（）和putc（）在没有任何特殊格式的情况下优化，现代编译器也可以优化
while（str [k]）！='\ 0'））{具有括号（'）'）
初始化您的变量str使用{}，这将把0分配给str的所有元素

更好地实现

我对此问题的实现是创建一个初始化的字符列表（256 max），然后添加1个到该列表中字符的ASCII值（来自str）。之后，打印那些值大于1的字符。

时间复杂度= o（n），其中n是字符串的长度
space complectity = o（no_of_chcharacters），其中no_of_of_characters是256

最终代码

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

static void print_dup(const char *str)
{
    size_t *count = calloc(1 << CHAR_BIT, sizeof(size_t));
    for(size_t i = 0; str[i]; i++)
    {
        count[(unsigned char)str[i]]++;
    }
    for (size_t i = 0; i < (1 << CHAR_BIT); i++)
    {
        if(count[i] > 1)
        {
            printf("`%c`, count = %zu\n", i, count[i]);
        }
    }
    free(count);
}

int main(void) {
    char str[51] = {};
    puts("Enter string:");
    if (scanf("%50s", str) != 1)
    {
        perror("bad input");
        return EXIT_FAILURE;
    }
    print_dup(str);
    return EXIT_SUCCESS;
}

Points to know

You don't need to pass the address of the variable str to scanf()
Don't use "%s", use "%<WIDTH>s", to avoid buffer-overflow
Always check whether scanf() conversion was successful or not, by checking its return value
Always use size_t to iterator over any array
i < strlen(str), makes the loop's time complexity O(n³), instead of O(n²), which also isn't very good you should check whether str[i] != 0. But, many modern compilers of C will optimize it by the way.
#include <Stdio.h> it is very wrong, stdio.h != Stdio.h
Call to printf() can be optimized using puts() and putc() without any special formatting, here also modern compiler can optimize it
while(str[k] != '\0')){ has a bracket (')')
Initialize your variable str using {}, this will assign 0 to all the elements of str

Better Implementation

My implementation for this problem is that create a list of character (256 max) with 0 initialized, and then add 1 to ASCII value of the character (from str) in that list. After that print those character whose value was greater than 1.

Time Complexity = O(n), where n is the length of the string
Space Complexity = O(NO_OF_CHARACTERS), where NO_OF_CHARACTERS is 256

Final Code

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

static void print_dup(const char *str)
{
    size_t *count = calloc(1 << CHAR_BIT, sizeof(size_t));
    for(size_t i = 0; str[i]; i++)
    {
        count[(unsigned char)str[i]]++;
    }
    for (size_t i = 0; i < (1 << CHAR_BIT); i++)
    {
        if(count[i] > 1)
        {
            printf("`%c`, count = %zu\n", i, count[i]);
        }
    }
    free(count);
}

int main(void) {
    char str[51] = {};
    puts("Enter string:");
    if (scanf("%50s", str) != 1)
    {
        perror("bad input");
        return EXIT_FAILURE;
    }
    print_dup(str);
    return EXIT_SUCCESS;
}

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