如何打印递增的“计数”？从这个函数变量？ [英] How can I print the incremented "count" variable from this function?

问题描述

  void  parse（ char  * str）
 {
  int  left =  0 ，right =  0 ，count =  0 ; 
  int  len = strlen（str）; 
 printf（ ％d，count）; 
 
  while （右< = len&& left< = right）{
  if （isDelimiter（str [right]）==  false ）
 right ++; 
 
  if （isDelimiter（str [right]）==  true & & left == right）{
  if （isOperator（str [right]）==  true ）{
 printf（ '％c'是操作员'\\ n ，str [right]）; 
 count = count + 1; 
} 
 
 right ++; 
 left = right; 
}  else   if （isDelimiter（str [right]）==  true && left！= right 
 ||（right == len&& left！= right））{
  char  * subStr = subString（str，left，right  -   1 ）; 
 
  if （isKeyword（subStr）==  true ）{
 printf（ '％s'是关键字\ n，subStr）; 
 count = count + 1; 
} 
 
  else   if （isInteger（subStr）==  true ）{
 printf（ '％ s'是一个INTEGER \ n，subStr）; 
 count = count + 1; 
} 
 
  else   if （isRealNumber（subStr）==  true ）{
 printf（ '％ s'是真实的数字\ n，subStr）; 
 count = count + 1; 
} 
 
  else   if （validIdentifier（subStr）==  true  
&& isDelimiter（str [right  -   1 ]）==  false ）{
 printf（ '％s'是一个有效的标识符\ n，subStr）; 
 count = count + 1; 
} 
 
  else   if （validIdentifier（subStr）==  false  
&& isDelimiter（str [right  -   1 ]）==  false ）{
 printf（ '％s'不是有效的标识符\ n，subStr）; 
 count = count + 1; 
} 
 left = right; 
} 
} 
  return ; 
} 
 
  //   DRIVER FUNCTION  
 < span class =code-keyword> int  main（）
 {
  //  此处字符串的最大长度为100  
  char  str [ 100 ] =   int a = b + 1c;; 
 
 parse（str）;  //  调用解析函数 
 
  return （ 0 ）; 
} 

我的尝试：



i尝试返回一个整数值，但该函数也有字符。

i需要递增的计数值。

我需要打印计数 每次执行循环时递增的变量

解决方案

有点不清楚你的意思，但是如果你想将字符添加到长度而不是添加每个字符为1或添加字符串长度。



您正确使用了strlen函数：

  char  str [ 100 ] =  < span class =code-string> int a = b + 1c;; 
  int  len = strlen（str）; 
  //  或添加到其他长度 
 totalLength + = len; 
 printf（ ％d，len）; 

void parse(char* str)
{
    int left = 0, right = 0, count=0;
    int len = strlen(str);
	printf("%d",count);
 
    while (right <= len && left <= right) {
        if (isDelimiter(str[right]) == false)
            right++;
 
        if (isDelimiter(str[right]) == true && left == right) {
            if (isOperator(str[right]) == true){
                printf("'%c' IS AN OPERATOR\n", str[right]);
				count=count+1;
			}
 
            right++;
            left = right;
        } else if (isDelimiter(str[right]) == true && left != right
                   || (right == len && left != right)) {
            char* subStr = subString(str, left, right - 1);
 
            if (isKeyword(subStr) == true){
                printf("'%s' IS A KEYWORD\n", subStr);
				count=count+1;
			}
 
            else if (isInteger(subStr) == true){
                printf("'%s' IS AN INTEGER\n", subStr);
				count=count+1;
			}
 
            else if (isRealNumber(subStr) == true){
                printf("'%s' IS A REAL NUMBER\n", subStr);
				count=count+1;
			}
 
            else if (validIdentifier(subStr) == true
                     && isDelimiter(str[right - 1]) == false){
                printf("'%s' IS A VALID IDENTIFIER\n", subStr);
				count=count+1;
			}
 
            else if (validIdentifier(subStr) == false
                     && isDelimiter(str[right - 1]) == false){
                printf("'%s' IS NOT A VALID IDENTIFIER\n", subStr);
				count=count+1;
			}
            left = right;
        }
    }
    return ;
}
 
// DRIVER FUNCTION
int main()
{
     // maximum legth of string is 100 here 
    char str[100] = "int a = b + 1c; ";
 
    parse(str); // calling the parse function
 
    return (0);
}

What I have tried:

i tried returning an integer value, but the function also has character.
i need the incremented value of the count.
I need to print the "count" variable that is incremented each time the loop is executed

解决方案

It is a bit unclear what you mean, but if you want to add the characters to the lenght than add 1 for every char or add the the string length.

You correctly used the strlen function:

char str[100] = "int a = b + 1c; ";
int len = strlen(str);
//or add to some other length
totalLength += len;
printf("%d", len);

这篇关于如何打印递增的“计数”？从这个函数变量？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！