C：如何比较两个字符串？ [英] C: How to compare two strings?

问题描述

编辑：这是一个重复的，我已经标记它是这样。参见[问题] 为什么＆QUOT; A＆QUOT; ！=＆QUOT; A＆QUOT;用C？

所以我想打印出视场一个结构中的特定信息。该字段包含字符串1。

每当我跑的printf（％S，record.fields [2]）; 输出 1 ;我没有格式的警告。

然而，当我检查方面对相应的字符串（在这种情况下，1），它失败的检查：

 如果（record.fields [2] ==1）{
    的printf（该字段为1！）;
}
 

解决方案

您需要使用的 STRNCMP 比较字符串：

 如果（STRNCMP（record.fields [2]，1，1）== 0）...
 

您需要比较为零，因为 STRCMP 返回零，当两个字符串是相同的。

不过，它看起来像你是不是比较字符串：相反，你正在寻找的字符串中特定的字符。在这种情况下，你需要使用一个字符常量，而不是字符串（单引号）：

 如果（record.fields [2] =='1'）...
 

Edit: This is a duplicate and I've flagged it as such. See [question] Why is "a" != "a" in C?

So I'm trying to print out a specific message depending on a field within a struct. The field contains the string "1".

Whenever I run printf("%s", record.fields[2]); the output is 1; I've no format warnings.

However, when I check the field against the corresponding string (in this case, "1"), it fails the check:

if (record.fields[2] == "1") {
    printf("The field is 1!");
}

解决方案

You need to use strncmp to compare strings:

if (strncmp(record.fields[2], "1", 1) == 0) ...

You need to compare to zero, because strcmp returns zero when two strings are identical.

However, it looks like you are not comparing strings: rather, you are looking for a specific character inside the string. In this case, you need to use a character constant instead of a string literal (with single quotes):

if (record.fields[2] == '1') ...

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