在C中将Char *转换为大写 [英] Converting Char * to Uppercase in C

问题描述

我正在尝试将c中的char *转换为大写，但是功能toupper()在这里不起作用.

I'm trying to convert a char * to uppercase in c, but the function toupper() doesn't work here.

我正在尝试获取temp值的名称，该名称是冒号之前的任何字符，在这种情况下为"Test"，然后我想将其完全大写.

I'm trying to get the name of the the value of temp, the name being anything before the colon, in this case it's "Test", and then I want to capitalize the name fully.

void func(char * temp) {
 // where temp is a char * containing the string "Test:Case1"
 char * name;

 name = strtok(temp,":");

 //convert it to uppercase

 name = toupper(name); //error here

}

我收到错误消息，该函数toupper期望一个int，但是接收到一个char *.问题是，我必须使用char *，因为这是函数要使用的((我不能在这里真正使用char数组，可以吗?).

I'm getting the error that the function toupper expects an int, but receives a char *. Thing is, I have to use char *'s, since that is what the function is taking in, (I can't really use char arrays here, can I?).

任何帮助将不胜感激.

推荐答案

toupper()转换单个char.

只需使用循环:

void func(char * temp) {
  char * name;
  name = strtok(temp,":");

  // Convert to upper case
  char *s = name;
  while (*s) {
    *s = toupper((unsigned char) *s);
    s++;
  }

}

详细信息:为所有unsigned char和EOF定义了标准库函数toupper(int).由于char可能已签名，请转换为unsigned char.

Detail: The standard Library function toupper(int) is defined for all unsigned char and EOF. Since char may be signed, convert to unsigned char.

某些操作系统支持执行此操作的函数: upstr() 和 strupr()

Some OS's support a function call that does this: upstr() and strupr()

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