为什么我不能在c中将字符串转换为int

Question

假设我们有一个char数组input = "12345678987654321"，为什么我不能用atoi()把它转换成整数12345678987654321？我已经尝试过了，它返回了一些我无法理解的数字。

代码如下：

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

int main() {
    char num[20];
    scanf("%s\n", num);
    printf("%d", atoi(num));
    return 0;
}

Answer 1

您得到奇怪输出的原因是输入的数字超出了int 类型可表示的值范围。 atoi() 函数不一定执行错误检查，如果数字不能表示为 int，它实际上具有未定义的行为。你应该改用strtol()。

您的程序还有另一个潜在的未定义行为：如果用户输入超过 19 个字符，scanf("%s\n", num) 可能会超出 num 数组的末尾，这会产生可怕的后果。您应该写scanf("%19s", num) 并测试返回值是否存在潜在的转换错误。

这是修改后的版本：

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

int main() {
    char num[20];

    if (scanf("%19s", num) != 1) {
        printf("no input\n");
        return 1;
    }
    int c = getchar();
    if (c != '\n' && c != EOF) {
        printf("too many characters\n");
        ungetc(c, stdin);
    }
    char *end;
    errno = 0;
    long value = strtol(num, &end, 10);
    if (end == num) {
        printf("input is not a number: %s", num);
    } else {
        if (errno != 0) {
            printf("number %s is too large for type long\n", num);
        } else
        if (value < INT_MIN || value > INT_MAX) {
            printf("number %ld is too large for type int\n", value);
        } else {
            printf("value is %d\n", (int)value);
        }
        if (*end != '\0') {
            printf("extra characters after the number: %s", end);
        }
    }
    return 0;
}

Answer 2

对于int，“12345678987654321”通常太大，atoi 不提供错误检测。

您可以使用strto… 函数将字符串中的数字转换为整数类型。下面是一个使用 strtoimax 的示例，它是 C 实现支持的最宽的有符号整数类型。还有其他类型的版本，例如strtoll for long long。

strto… 函数提供错误检查功能，因此您可以检测数字何时过大并将其报告给用户或以其他方式处理。

#include <ctype.h>      // For isdigit and isspace.
#include <errno.h>      // For errno.
#include <inttypes.h>   // For stroimax and PRIdMAX.
#include <stdint.h>     // For intmax_t.
#include <stdio.h>      // For printf, fprintf, getchar, and ungetc.
#include <stdlib.h>     // For exit.


int main(void)
{
    // Read string.
    char string[20];
    int n = 0; // Number of characters read so far.
    while (1)
    {
        int c = getchar();

        // When we see the end of the input stream, leave the loop.
        if (c == EOF)
            break;

        /* When we see a white-space character (including new-line), put it
           back into the stream and leave the loop.
        */
        if (isspace(c))
        {
            ungetc(c, stdin);
            break;
        }

        /* If we see character that is not a space or a digit, complain, unless
           it is the first character and is a sign, "+" or "-".
        */
        if (!isdigit(c) && !(n == 0 && (c == '+' || c == '-')))
        {
            fprintf(stderr, "Error, non-digit seen:  \"%c\".\n", c);
            exit(EXIT_FAILURE);
        }

        // Record the digit.
        string[n++] = c;

        if (sizeof string / sizeof *string <= n)
        {
            fprintf(stderr, "Error, too many digits, can handle only %zu.\n",
                sizeof string / sizeof *string - 1);
                /* ("sizeof *string" is one for ordinary "char" arrays, but
                   using it allows changing to wide character types if
                   desired.)
                */
            exit(EXIT_FAILURE);
        }
    };
    // Finish string with a null character.
    string[n] = '\0';

    // Attempt to convert string to an intmax_t.
    char *end;
    errno = 0; // Initialize error indicator.
    intmax_t number = strtoimax(string, &end, 0);

    /* Test whether the string failed to match the form for a numeral, because
       it was empty or was just a sign with no digits.
    */
    if (end == string)
    {
        fprintf(stderr, "Error, numeral has no digits.\n");
        exit(EXIT_FAILURE);
    }

    // Check for error.
    if (errno != 0)
    {
        fprintf(stderr, "Error, number is too large for intmax_t.\n");
        exit(EXIT_FAILURE);
    }

    /* Test whether the whole string was used in the conversion. It should have
       been, considering we ensured each character was a digit, but a check
       like this can be useful in other situations, where we are parsing a
       numeral from a more general string.
    */
    if (*end != 0)
    {
        fprintf(stderr, "Error, unexpected character in numeral, \"%c\".\n",
            *end);
        exit(EXIT_FAILURE);
    }

    // Print the number.
    printf("The number is %" PRIdMAX ".\n", number);
}

Answer 3

您输入的“12345678987654321” atoi 在 int 中，而 int 的范围来自 -2,147,483,648 - 2,147,483,647 所以为了修复它使用 atol() l-long，这只是因为范围或者如果您使用的是 32 位机器那么你需要使用 long long int 即 atoll() 而不是 atol() data size

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
    char num[20];
    scanf("%s",num);
    //printf("\n%s",num);
    printf("\n%ld",atol(num)); //<- atoi(num) to atol(num)
    return 0;
}

atol()查看更多atoi()、atol()和atoll()