如何检查C中的字符串是否为空

I have a void function which take the string from a global variable and prints them.

我有一个void函数，它从全局变量中取出字符串并打印出来。

The problem is that the void function prints a message/does stuff even if the string is empy. so if the global variable is empt it just prints "message is:" i only want to do something if the string has characters.

问题是，void函数打印消息/做事情，即使字符串是空的。因此，如果全局变量是空的，它只打印“message is:”如果字符串有字符，我只想做一些事情。

what i want to do is to check if the string is not empty, then print a message.

我要做的是检查字符串是否为空，然后打印一条消息。

// global variable

char msg[30];
scanf("%s", &msg);

void timer(void) {
    //i want to only print if there is a message
    printf("message: %s", msg);
}

thanks for any help

感谢任何帮助

5 个解决方案

#1

The easy way to do it would be like this:

简单的方法是:

if (msg[0])
    printf("message: %s", msg);

If, at some later date, msg is a pointer, you would first want to assure it's not a NULL pointer.

如果在以后的某个时候，msg是一个指针，那么您首先要确保它不是一个空指针。

if (msg && msg[0])
    printf("message: %s", msg);

A test program to demonstrate:

测试程序演示:

#include <stdio.h>
#include <stdlib.h>

char *msg;

void test() {
    if (msg && msg[0])
        printf("string is %s\n", msg);
}

int main()
{
    msg = NULL;
    test();
    msg = calloc(10, 1);
    test();
    msg[0] = 'm';
    msg[1] = 'e';
    test();
    free(msg);
}

On my machine the output is:

在我的机器上输出是:

string is me

字符串是我

#2

Strings in C are represented as character arrays, terminated by a character whose value is zero (often written as '\0'). The length of a string is the number of characters that come before the zero.

C中的字符串表示为字符数组，终止符的值为0(通常写成'\0')。字符串的长度是在0之前的字符数。

So, a string is empty if the first character in the array is zero, since then by definition no characters came before the zero.

因此，如果数组中的第一个字符是0，那么字符串就是空的，因为根据定义，在0之前没有任何字符。

This can be checked like so:

可以这样检查:

if(msg[0] != '\0')
{
  /* string isn't empty! */
}

This is very explicit code, the shorter if(msg[0]) or if(*msg) has the exact same semantics but can be slightly harder to read. It's mostly a matter of personal style.

这是非常明确的代码，较短的if(msg[0])或if(*msg)具有完全相同的语义，但读起来可能会稍微困难一些。这主要是个人风格的问题。

Note though that in your case, when you use scanf() to read into msg, you should check the return value from scanf() before checking the contents of msg. It's perfectly possible for scanf() to fail, and then you can't rely on msg being set to an empty string.

请注意，在您的例子中，当您使用scanf()读取msg时，您应该在检查msg的内容之前检查scanf()的返回值。scanf()完全有可能失败，然后您不能指望msg被设置为空字符串。

#3

This should work for you:

这应该对你有用:

The Message will only be printed if it's longer then 0.

消息只有在比0长时才会被打印出来。

#include <stdio.h>
#include <string.h>

void timer() {

    char msg[30];
    scanf(" %s", msg);

    if(strlen(msg) != 0)
        printf("message: %s", msg);
}

int main() {

    timer();

    return 0; 
}

#4

Here are all the ways I found (and can think of) to check if a string is empty. Sorted by efficiency.

下面是我找到(并能想到的)检查字符串是否为空的所有方法。按效率。

msg[0] == '\0'
味精[0]= = ' \ 0 '
strlen(msg) == 0
strlen(味精)= = 0
!strcmp(msg, "")
! strcmp(味精”、“)

And this, which should only happen if the user somehow avoided input (e.g. with Ctrl+Z) or there was an error

只有当用户以某种方式避免输入(例如使用Ctrl+Z)或出现错误时才会发生这种情况

scanf("%s", &msg) <= 0
scanf(“% s”,味精)< = 0

#5

-1

Ok,you can try this:

好吧,你可以试试这个:

char msg[30];
scanf("%s",&msg);

void timer(void)
{
    //i want to only print if there is a message
    if(!msg.length == 0 )
        printf("message: %s", msg);
}

”

#1