Question

我遇到了一个我真的不知道如何解决的问题。我主要使用Python编写代码，这是我的第一个C语言程序。

#include <stdio.h>

int ask(void) {
  scanf("Var");
  return 0;
}

int count(ask) {
  scanf("number1");
  return 0;
}

int main(void) {
  printf("This is my first program!\n");
  printf("I hope this program turns out well.");
  printf("I don't really know what to do, but i think im progressing.\n");
  printf("But yeah, This is my first program.\n");
  printf("Type an Number");
  ask();
  count(ask());
  printf("Thanks!");
  printf(%count%);
  return 0;
}

但是，我不断收到错误消息。

main.c:22:10: error: expected expression
  printf(%count%);
         ^
main.c:22:17: error: expected expression
  printf(%count%);
                ^
2 errors generated.
compiler exit status 1

我想要它做的是，用户键入一个数字，然后打印出该数字。虽然还不完整。我要它写数字1-用户输入，当数字正确时，它会显示“您的数字是：”（数字）

Answer 1

问题（正如已经指出的那样）是您实际上并没有从scanf（）调用中获取和存储值。此外，printf(%count%)不是有效的C语法。您需要使用printf("%d", count)。

将所有内容放在一起：

#include <stdio.h>

int ask(void) {
  int input_number;
  scanf("%d", &input_number);
  getchar();  # This is so that the '\n' in is read when you hit Enter
  return input_number;
}

int main(void) {
  printf("This is my first program!\n");
  printf("I hope this program turns out well.");
  printf("I don't really know what to do, but i think im progressing.\n");
  printf("But yeah, This is my first program.\n");
  printf("Type an Number");
  int input_number = ask();
  printf("Thanks!");
  printf("The number you entered is %d\n", input_number);
  return 0;
}

为避免犯此类错误，需要阅读一些内容：

printf教程：https://www.tutorialspoint.com/c_standard_library/c_function_printf.htm

scanf教程：https://www.tutorialspoint.com/c_standard_library/c_function_scanf.htm

Answer 2

您使用了不正确的格式进行打印。您正在使用此代码 printf(%count%);，而应使用此代码printf("%d",count(variablename));

这是您可以使用的代码：

#include <stdio.h>

 int ask(void) {
    //scanf("Var");
    int var; //declaring integer type varible in C
    scanf("%d",&var);  //taking input

    return var; //returning what is being input by user
 }

//int count(ask) //its function you are passing
//so it should have parenthesis
int count( int a ) 
{

      return a;

 }

  int main(void) {
   printf("This is my first program!\n");
   printf("I hope this program turns out well.");
   printf("I don't really know what to do, but i think im progressing.\n");
   printf("But yeah, This is my first program.\n");
   printf("Type an Number: ");

           //ask();   
    count(ask()); // when you will call it will automatically run ask as well 
           // and it will pass int value that user will input 
    printf("Thanks!");
    int var1=ask();   // this varible var1 will store value returned by ask() 
    printf("%d",count(var1) );  //this will print that value

                 //I dont know what you wanted to do here
                 //maybe you wanted to print value returned by count function
                 //so it can be done by this
   return 0;
}

printf格式：printf("%formatspecifier",variable name);的格式说明符是%d的整数。 %f为浮动值。 %c作为字符。 s表示字符串，依此类推。

Answer 3

在C语言中，我们使用%指定输出/输入类型。％d代表整数，％f代表浮点数，％c代表字符，％s代表字符串，这就是您的基本知识。

对于printf：

printf("%d", varname);

对于scanf：

scanf("%d", &varname);

'＆'表示在内存中的位置。

您的程序遇到很多语法错误，下面是一些代码：

#include<stdio.h>

int ask(){
    
    int varin;
    
    scanf("%d", &varin);
    return (varin);
}

int count(int countin){
    return (countin);   //just an example code, or whatever you wanna do here.
}

int main(){
    
    int out;
    
    printf("Whatever\n");
    out = ask();
    count(out);
    printf("%d", out);
    
    return 0;
}

C中的编译器错误，错误：预期表达式

3 个答案: