Question

我有一个程序可以将char整数转换为实际的整数。它必须与教授编写的Main方法兼容。我想我有一个实际转换的解决方案，但我不断得到一个无限循环。我知道我应该回归Null，但我不知道如何。请任何人可以帮忙吗？此函数也应该忽略char字母，并在每个字母后面都有一个新空格。因此123fgf456将打印123（换行符）456。

The function I need help with.

    #include <stddef.h>
    /*
     * Scans inputString, ignoring leading whitespace (spaces, tabs, and newlines)
     * to find the first decimal digit, which it interprets as the most significant
     * digit of a decimal number, and continues scanning until finding the first
     * non-decimal digit. The digits found are converted to an integer, which is
     * stored in the location pointed to by integerPtr. This function returns a
     * pointer to the first non-digit after the first digit, unless a
     * non-whitespace, non-digit is encountered before a digit, in which case,
     * NULL is returned and the location pointed to by integerPtr is not changed.
     *
     * @param integerPtr A pointer to the integer in which to store the integer
     *        converted from the ASCII string
     * @return a pointer to the first non-digit character found if a number was
     *        successfully converted, NULL if not
     */

   char * asciiToInteger(char *inputString, int *integerPtr) {
    int i =0; int j=0; int num =0; char terminate; 
    if(inputString[i] == '\0') return NULL;
    if(inputString[i] == ' '){return NULL;}
    while(*inputString != '\0')
    {   

        if(*inputString >= '0' && *inputString <= '9')
        {
            if(inputString[i] == ' '){break;}
            num = num *10 + inputString[i]- '0';
            *integerPtr = num;
        }
        inputString++;

    }

    return inputString;
}




    #include <stdio.h>
    #include <stddef.h> // for definition of NULL

    char * asciiToInteger(char inputString[], int *integerPtr);
    int main() {
       char inputBuffer[1024];
       char *ptr = NULL;
       int integer = 8888;
       while(fgets(inputBuffer, sizeof(inputBuffer), stdin)) {
          ptr = inputBuffer;
          int done = 0;
          while(!done) {
             char *newPtr = asciiToInteger(ptr, &integer);
             if(newPtr == NULL) {
                if(*ptr != '\0')
                   ++ptr; // Skip over offending character
                else
                   done = 1;
             } else {
                printf("%d\n", integer);
            ptr = newPtr;
             }
          }
       }

       return 0;
    }

Answer 1

char * asciiToInteger(char *inputString, int *integerPtr) {
  int n = 0;
  int index = 0;
  while(*inputString != NULL) {
    if (*inputString >= '0' && *inputString <= '9') {
      n = n * 10 + *inputString - '0';
      ++inputString;
      integerPtr = &n;
      index += 1;
    } else if(*inputString < '0' || *inputString > '9') {
        if(*integerPtr) std::cout << *integerPtr << std::endl;
        n = 0;
        ++inputString;
    } else {
        if (*integerPtr) std::cout << *integerPtr << std::endl;
        ++inputString;
    }
}
  if(*integerPtr) std::cout << *integerPtr << std::endl;
  return inputString;
}

以上是C ++实现，这是使用printf的实现：

char * asciiToInteger(char *inputString, int *integerPtr) {
  int n = 0;
  int index = 0;
  while(*inputString != NULL) {
    if (*inputString >= '0' && *inputString <= '9') {
      n = n * 10 + *inputString - '0';
      ++inputString;
      integerPtr = &n;
      index += 1;
    } else if(*inputString < '0' || *inputString > '9') {
        if(*integerPtr) printf("%d\n", *integerPtr);
        n = 0;
        ++inputString;
    } else {
        if (*integerPtr) printf("%d\n", *integerPtr);
        ++inputString;
    }
  }
  if(*integerPtr) printf("%d\n", *integerPtr);
  return inputString;
}

我发现你的提示相当模糊，但这应该通过那些测试用例。现在，还要求您在初始数字后返回指向第一个非数字的指针。您可以返回＆amp; inputString [index - 1]，但是，您不能提前返回。

Answer 2

你需要一个标记来记住你是否看过任何数字。然后，如果您看到一个非数字，但您还没有看到任何数字，请立即使用browserHistory进行纾困。

硬件帮助今晚到期。 Char To Int（帮助返回空）

2 个答案: