如何修复错误格式，指定类型为"char *"，但参数为类型"char" [英] How to fix error Format specifies type 'char *' but the argument has type 'char'

问题描述

我收到警告说: 对于学生变量，格式指定类型为'char *'，但参数的类型为'char'".我正在将代码从一本书中复制/粘贴到xcode中，并且不确定如何解决此问题.控制台中唯一显示的内容是(lldb)".任何建议

I get a warning saying: "Format specifies type 'char *' but the argument has type 'char'" for the student variable. I am copying/pasting the code out of a book into xcode and am not sure how to fix this. The only thing that prints in the console is "(lldb)". Any advice

#include <stdio.h>

void congratulateStudent(char student, char course, int numDays)
{
    printf("%s has done as much %s Programming as I could fit into %d days.\n", student, course, numDays);
}

int main(int argc, const char * argv[])
{
    // insert code here...
    congratulateStudent("mark", "Cocoa", 5);
    return 0;
}

推荐答案

void congratulateStudent(char *student, char *course, int numDays)

%s表示您将要打印一个字符串(字符数组)

the %s means that you are going to print a string ( array of chars)

和char student这表示学生是char类型

所以这里的学生不是指向字符串的指针

so student here is not a pointer to a string

要将学生类型从char更改为字符串指针，您必须向学生char *student

In order to change the student type from char to a string pointer you have to add asterisk to student char *student

在您的代码中，您正在使用输入参数字符串"mark"调用congratulateStudent.因此，要支持此字符串，应将输入参数student定义为字符串的指针

In your code you are calling the congratulateStudent with input parameter string "mark". So to support this string the input parameter student should be defined as pointer of string

所以您在学生的定义中缺少星号

so you are missing the asterisk in the definition of student

与course

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