C语言 指针和整数的警告比较
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Warning comparison between pointer and integer
提问by catee
I am getting a warning when I iterate through the character pointer and check when the pointer reaches the null terminator.
当我遍历字符指针并检查指针何时到达空终止符时，我收到一条警告。
const char* message = "hi";
//I then loop through the message and I get an error in the below if statement.
if (*message == "warning: comparison between pointer and integer
('int' and 'char *')
") {
...//do something
}
The error I am getting is:
我得到的错误是：
if (*message == 'if (*message == "warning: comparison between pointer and integer
('int' and 'char *')
") {
')
I thought that the *in front of messagedereferences message, so I get the value of where message points to? I do not want to use the library function strcmpby the way.
我认为*前面的message取消引用消息，所以我得到消息指向的值？strcmp顺便说一下，我不想使用库函数。
回答by blakelead
It should be
它应该是
if(*message == '##代码##') ...
if(message[0] == '##代码##') ...
if(!*message) ...
In C, simple quotes delimit a single character whereas double quotes are for strings.
在 C 中，单引号用于分隔单个字符，而双引号用于字符串。
回答by dbush
This: "\0"is a string, not a character. A character uses single quotes, like '\0'.
this:"\0"是一个字符串，而不是一个字符。字符使用单引号，例如'\0'.
回答by Fiddling Bits
In this line ...
在这一行...
##代码##... as you can see in the warning ...
......正如你在警告中看到的......
##代码##... you are actually comparing an intwith a char *, or more specifically, an intwith an address to a char.
...您实际上是将 anint与 a进行比较char *，或者更具体地说，将 anint与一个地址进行比较char。
To fix this, use one of the following:
要解决此问题，请使用以下方法之一：
##代码##On a side note, if you'd like to compare strings you should use strcmpor strncmp, found in string.h.
附带说明一下，如果您想比较字符串，您应该使用strcmp或strncmp，在string.h.
