如果测试的整数使用位操作大写ASCII字母

问题描述

有关的转让，我试图让一些C code，它仅使用位操作来测试是否一个整数是ASCII大写字母。这封信将其ASCII code以整数形式给出，这意味着的0x41＆LT; = ourint＆LT; = 5AH即可。做一些研究后，我了解到，小写和大写字母之间的差异是这个词的二进制版本的第六位。大写有一个 1 ，下有一个 0 。

For an assignment, I'm trying to make some code in C that uses only bit manipulation to test if an integer is an ASCII uppercase letter. The letter will be given by its ASCII code in integer form, meaning 0x41 <= ourint <= 0x5A. After doing some research I learned that the difference between a lower case and an upper case letter was the sixth digit of the binary version of the word. Uppercase had a 1 and lower had a 0.

我几乎有它，但我的code将无法工作。

I am almost have it, but my code won't work.

我至今是

!(((x>>6)<<31) & 0)

它的作用是转移的code，以获得第六位作为第一数量的权利6，然后向右移动31让它成为全部为0或1，其次是31 0。我必须让这个如果是大写字母则返回1，所以我用感叹号来表示。

What it does is shifts the code to the right 6 to get the sixth digit as the first number, then shifts right 31 to get it to be either all 0's or 1 followed by 31 0's. I have to make it so that if it is uppercase it returns 1, so I use the exclamation point to that.

编辑：我的新的code是

!((~(((x & 32)>>5))<<31))>>31)

但现在我被困在为0x7FFFFFFF

推荐答案

您可以测试是否ASCII字母 C 是大写通过检查为0x20 位，它必须是 0 大写和 1 小写：

You can test if an ASCII letter c is upper case by checking its 0x20 bit, it must be 0 for uppercase and 1 for lowercase:

if (!(c & 0x20))
    printf("ASCII letter %c is uppercase\n", c);

但要注意，如果你不已经知道， C 是一个字母这个测试是行不通的。它会错误匹配'@'和'['，'\\\\'，]，'^'和'_'，并与高位192设置为223个字符，这是不ASCII的一部分，但都是有效的 unsigned char型值的整个范围。

but be aware that this test does not work if you don't already know that c is a letter. It would erroneously match '@' and '[', '\\', ']', '^' and '_', and the whole range of characters with the high bit set from 192 to 223, which are not part of ASCII but are valid unsigned char values.

如果你想有一个单独的测试，如果来验证ç是大写ASCII字母，请尝试：

If you want a single test to verify if c is an uppercase ASCII letter, try:

if ((unsigned)(c - 'A') <= (unsigned)('Z' - 'A'))
     printf("%c is an uppercase ASCII letter\n", c);

编辑：目前还不清楚您的我不允许使​​用if语句，或任何种类的类型转换操作的意思。我必须测试，看看是否数字是两个数字之间，包括远远超出ASCII code的范围内的数字，并返回1，如果是要不0 的

• 如果您知道 C 是一个字母，既！（C＆安培;为0x20）和（（（C＆GT;＆GT; 5）及1）^ 1）将值 1 如果 ç是大写和 0 如果没有


• 如果 C 可以是任何整数值，只是写正规的比较（C＆GT; ='A'和;和C＆LT; ='Z'），编译器将通过尝试危险的位变换花样产生更好的code比你。

• If you know c is a letter, both !(c & 0x20) and (((c >> 5) & 1) ^ 1) will have value 1 if c is uppercase and 0 if not.
• If c can be any integer value, just write the regular comparison (c >= 'A' && c <= 'Z') and the compiler will produce better code than you would by attempting hazardous bit-twiddling tricks.

再次编辑

由于 C 可以是任何整数值，你只被允许位操作，这里是另一种解决方案：（（C＆GT;＆GT; 5）^ 2）及（0x07fffffeU＆GT;＆GT;（三及31条））。下面是一个程序来测试这一点：

Since c can be any integer value and you are only allowed bit manipulations, here is another solution: !((c >> 5) ^ 2) & (0x07fffffeU >> (c & 31)). Below is a program to test this:

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

static int uppertest(int c) {
    return !((c >> 5) ^ 2) & (0x07fffffeU >> (c & 31));
}

int main(int argc, char *argv[]) {
    for (int i = 1; i < argc; i++) {
        int c = strtol(argv[i], NULL, 0);
        printf("uppertest(%d) -> %d\n", c, uppertest(c));
    }
    return 0;
}

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