Question

我有一个练习要在MIPS程序集中解决（我有一些疑问，但其他事情很清楚），但是编写代码时遇到了一些问题。练习问我：编写一个程序，从键盘获取一个字符串，计算具有较多出现次数的字符的出现次数并显示它。

我如何检查所有26个字符，找出谁有较高的出现次数？

实施例：给我一个字符串：Hello world！出现次数较高的字符是：l

非常感谢将来的回答。

P.S。这是我的第一部分：

        #First message      
        li $v0, 4
        la $a0, mess
        syscall

        #Stack space allocated
        addi $sp, $sp, -257 

        #Read the string
        move $a0, $sp
        li $a1, 257
        li $v0, 8
        syscall

Answer 1

由于这是你的任务，我将把MIPS程序集实现留给你。我将以更高级别的语言向您展示代码的逻辑：

// You'd keep these variables in some MIPS registers of your choice
int c, i, count, max_count=0;
char max_char;

// Iterate over all ASCII character codes   
for (c = 0; c < 128; c+=1) {
    count = 0;
    // Count the number of occurences of this character in the string
    for (i = 0; string[i]!=0; i+=1) {
        if (string[i] == c) count++;
    }
    // Was is greater than the current max?
    if (count > max_count) {
        max_count = count;
        max_char = c;
    }
}

// max_char now hold the ASCII code of the character with the highest number
// of occurences, and max_count hold the number of times that character was
// found in the string.

Answer 2

@Michael，在我发布之前，我看到你回答了，我只想用更详细的答案重复一下。如果您编辑自己的内容以添加更多解释，那么我将删除我的。我没有直接编辑你的，因为你发布的时候我已经在中途了。无论如何：

@Marco：
您可以创建一个包含26个计数器的临时数组（初始化为0）。

每个计数器对应于每个字母（即每个字母出现的数字）。例如，counter[0]对应于字母＆＃39;＆＃39;，counter[1]出现的字母＆＃39; b＆＃39;等等...

然后迭代输入字符序列中的每个字符，并为每个字符执行：
a）获取counter数组中字符的索引 b）将counter["obtained index"]增加1.

要获取角色的索引，您可以执行以下操作：
a）首先要确保角色不是资本，即只有＆＃39; a＆＃39;到＆＃39; z＆＃39;允许而不是＆＃39; A＆＃39;到＆＃39; Z＆＃39;如果不是，请转换它 b）提取字母＆＃39; a＆＃39;从角色。通过这种方式＆＃39; a＆＃39; a＆＃39;得到0，＆＃39; b＆＃39; - ＆＃39; a＆＃39;给出1，＆＃39; c＆＃39;＆＃39; a＆＃39;给2等等......

我将使用C语言进行演示，因为它是您在MIPS上的练习（我的意思是目标是学习MIPS汇编语言）：

#include <stdio.h>

int main()
{
    //Maximum length of string:
    int stringMaxLength = 100;

    //Create string in stack. Size of string is length+1 to
    //allow the '\0' character to mark the end of the string.
    char str[stringMaxLength + 1];

    //Read a string of maximum stringMaxLength characters:
    puts("Enter string:");
    scanf("%*s", stringMaxLength, str);
    fflush(stdin);

    //Create array of counters in stack:
    int counter[26];

    //Initialize the counters to 0:
    int i;
    for (i=0; i<26; ++i)
        counter[i] = 0;

    //Main counting loop:
    for (i=0; str[i] != '\0'; ++i)
    {
        char tmp = str[i]; //Storing of str[i] in tmp, to write tmp if needed,
        //instead of writing str[i] itself. Optional operation in this particular case.

        if (tmp >= 'A' && tmp <= 'Z') //If the current character is upper:
            tmp = tmp + 32; //Convert the character to lower.

        if (tmp >= 'a' && tmp <='z') //If the character is a lower letter:
        {
            //Obtain the index of the letter in the array:
            int index = tmp - 'a';

            //Increment its counter by 1:
            counter[index] = counter[index] + 1;
        }
        //Else if the chacacter is not a lower letter by now, we ignore it,
        //or we could inform the user, for example, or we could ignore the
        //whole string itself as invalid..
    }

    //Now find the maximum occurences of a letter:
    int indexOfMaxCount = 0;
    int maxCount = counter[0];
    for (i=1; i<26; ++i)
        if (counter[i] > maxCount)
        {
            maxCount = counter[i];
            indexOfMaxCount = i;
        }

    //Convert the indexOfMaxCount back to the character it corresponds to:
    char maxChar = 'a' + indexOfMaxCount;

    //Inform the user of the letter with maximum occurences:
    printf("Maximum %d occurences for letter '%c'.\n", maxCount, maxChar);

    return 0;
}

如果您不明白为什么我通过添加32将大写字母转换为较低的字母，请继续阅读：

每个字符对应于内存中的整数值，当您对字符进行算术运算时，就像在编码表中将它们设置为相应的数字一样。
编码只是一个将这些字母与数字相匹配的表格例如＆＃39; a＆＃39;对应于ASCII encoding/decoding/table中的数字97 例如＆＃39; b＆＃39;对应ASCII encoding/decoding/table中的数字98。

所以＆＃39; +1给出97 + 1 = 98这是字符＆＃39; b＆＃39;。它们都是内存中的数字，区别在于你如何表示它们（ decode ）。当然，编码的同一表也用于解码。

示例：

printf("%c", 'a'); //Prints 'a'.  
printf("%d", (int) 'a'); //Prints '97'.  
printf("%c", (char) 97); //Prints 'a'.  
printf("%d", 97); //Prints '97'.  
printf("%d", (int) 'b'); //Prints '98'.  
printf("%c", (char) (97 + 1)); //Prints 'b'.  
printf("%c", (char) ( ((int) 'a') + 1 ) ); //Prints 'b'.  
//Etc...  
//All the casting in the above examples is just for demonstration,  
//it would work without them also, in this case.

MIPS，位于堆栈中的字符串中出现的次数

2 个答案: