如何在C中生成期望值

Question

我是 C 语言的新手，我正在尝试编写一个简单的自助银行控制台程序。我想做一个生成帐号的函数，所以我写了下面的代码：

int first = 7856123490;
int accountNumber;

int generateAccNo ()
{
    printf("%d", first);
    first = first + 1;
    accountNumber = first;
    return (accountNumber);
}

这里的整个想法是，当我调用这个函数时，它将获取名为 first 的变量中的值并将其加 1 并替换 first 和 [=24= 的值] 该值为 accountNumber.

正如我所期望的，这个函数第一次运行它应该 return 7856123491 这个值，第二次它应该 return 7856123492。那我该怎么做呢？

Answer 1

long int first = 7856123490;
long int accountNumber;

int generateAccNo ()
{
    printf("%ld \n", first);
    first = first + 1;
    accountNumber = first;
    return (accountNumber);
}

int 的尺寸很小，无法容纳如此大的数字，因此我们将在声明 int

long

为了在将 long int 与 %d 一起使用时抑制警告，我们将对 long int

%ld

Answer 2

更便携的解决方案：在中使用定长类型。

#include <inttypes.h>
#include <stdio.h>

int64_t first = 7856123490;
int64_t accountNumber;

int64_t generateAccNo() {
  first = first + 1;
  accountNumber = first;
  return (accountNumber);
}

#define TEST_CNT 10
int main(int argc, char **argv) {
  for (int64_t i = 0; i < TEST_CNT; ++i) {
    int64_t res = generateAccNo();
    printf("%" PRId64 " %016" PRIX64 "\n", res, (uint64_t) res);
  }

  return 0;
}

Answer 3

使用大数字的简单方法，即使您有 N 位数的银行帐户也可以使用

void big_number_inc(char * const big_number, const int size) {
    int i = size - 1;

    big_number[i]++;
    while ((i >= 0) && (big_number[i] > ASCII_9)) {
        big_number[i--] = ASCII_0;
        if (i >= 0) big_number[i]++;
    }
}

完整参考。代码：

#include <stdio.h>

#define ARRAY_SIZE(x)       (sizeof(x)/sizeof(x[0]))
#define ASCII_0             48      // '0' assci value
#define ASCII_9             57      // '9' assci value

// internal use (debug only)
static void inline big_number_print(const char *const big_number, int size) {
    for (int i = 0; i < size; i++)
        printf("%c ", big_number[i]);
    printf("\n");
}

void big_number_inc(char * const big_number, const int size) {
    int i = size - 1;

    big_number[i]++;
    while ((i >= 0) && (big_number[i] > ASCII_9)) {
        big_number[i--] = ASCII_0;
        if (i >= 0) big_number[i]++;
    }
}

int main(int argc, char *argv[]) {
    // note: if max length of number doesn't match initial string length - 1, 
    // fill remaining with '0' or revert and start working from offset 0
    char number[] = { '0', '0', '7', '8', '5', '6', '1', '2', '3', '4', '9', '0' };

    for (int i = 0; i < 20; i++) {
        big_number_inc(number, ARRAY_SIZE(number));

        // wanna see the result
        big_number_print(number, ARRAY_SIZE(number));
    }

    return 0;
}

如何在C中生成期望值

how to generate expected value in C

c

console-application