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如何限制c中的输入

发布于 2024-08-10 21:20:50 字数 58 浏览 11 评论 0原文

我想阻止我的程序接受除 int 之外的任何其他类型的输入。如何检查输入的类型而不将其分配给变量？在C中

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堇色安年 2024-08-17 21:20:50

请参阅 strtod、strtol 和朋友们。

所有这些函数都带有一个输出参数（通常称为endptr），该参数显示转换结束的位置。您可以使用该信息来确定要转换的输入是整数、浮点数还是根本不是数字。

该策略是尝试将输入解析为以 10 为基数的长度。如果这不起作用（即如果存在未转换的字符），请查看将其解析为双精度是否有效。如果两者都不起作用，则输入不是数字。显然，您必须决定要对数字使用哪种基本类型。您可以进行更多检查和改进，但这里是一个简单的示例。

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

struct a_number {
    unsigned char is_long;
    union {
        double d;
        long i;
    };
};

int main(void) {
    int n;
    char *input_strings[3] = {
        "1234","1234.56", "12asdf"
    };

    struct a_number *numbers[3];

    for (n = 0; n < 3; ++n) {
        char *start = input_strings[n];
        char *end;
        long i = strtol(start, &end, 10);
        if ( *end == '\0' ) {
            struct a_number *num = malloc(sizeof(*num));
            assert( num );
            num->is_long = 1;
            num->i = i;
            numbers[n] = num;
        }
        else {
            double d = strtod(start, &end);
            if ( *end == '\0' ) {
                struct a_number *num = malloc(sizeof(*num));
                assert( num );
                num->is_long = 0;
                num->d = d;
                numbers[n] = num;
            }
            else {
                numbers[n] = NULL;
            }
        }
    }

    for (n = 0; n < 3; ++n) {
        if ( numbers[n] ) {
            if ( numbers[n]->is_long ) {
                printf("%ld\n", numbers[n]->i );
            }
            else {
                printf("%g\n", numbers[n]->d );
            }
        }
    }

    return 0;
}

See strtod, strtol and friends.

All of those function take an out parameter (generally referred to as endptr) which shows you where the conversion ended. You can use that information to decide if the input you wanted to convert was an integer or floating point number or not a number at all.

The strategy is to try to parse the input as a base 10 long. If that does not work (i.e. if there were unconverted characters), see if parsing it as a double works. If neither works, the input is not a number. Clearly, you would have to decide what basic type you would want to use for numbers. You can build in more checks and refinements, but here is a simple example.

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

struct a_number {
    unsigned char is_long;
    union {
        double d;
        long i;
    };
};

int main(void) {
    int n;
    char *input_strings[3] = {
        "1234","1234.56", "12asdf"
    };

    struct a_number *numbers[3];

    for (n = 0; n < 3; ++n) {
        char *start = input_strings[n];
        char *end;
        long i = strtol(start, &end, 10);
        if ( *end == '\0' ) {
            struct a_number *num = malloc(sizeof(*num));
            assert( num );
            num->is_long = 1;
            num->i = i;
            numbers[n] = num;
        }
        else {
            double d = strtod(start, &end);
            if ( *end == '\0' ) {
                struct a_number *num = malloc(sizeof(*num));
                assert( num );
                num->is_long = 0;
                num->d = d;
                numbers[n] = num;
            }
            else {
                numbers[n] = NULL;
            }
        }
    }

    for (n = 0; n < 3; ++n) {
        if ( numbers[n] ) {
            if ( numbers[n]->is_long ) {
                printf("%ld\n", numbers[n]->i );
            }
            else {
                printf("%g\n", numbers[n]->d );
            }
        }
    }

    return 0;
}

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