允许简单的if语句,而不是在codestyle中没有大括号

Question

我使用checkstyle来检查我的java代码是否符合我们项目的指导原则.

但是,我们有一个指南,我无法弄清楚如何使用此工具进行检查.我们想要允许简单的if(理解是否没有其他,没有其他条件结构)没有大括号,就像在这个例子中:

// valid
if(condition) callFunction();

// invalid
if(condition) for(int i = 0; i < someValue; i++) callFunction(i);

// valid
if(condition) {
    for(int i = 0; i < someValue; i++) {
        callFunction(i);
    }
}

// invalid
if(condition) callFunction();
else callOtherFunction();

可以讨论这个约定,但这是我们选择的.它允许减少非常琐碎的if语法,但确保我们对更复杂的结构有良好的缩进和块定界.

任何帮助都将非常感激.

我还准备做一些代码来执行此检查,如果没有可用的,但实际上不知道从哪里开始.在最后一次报道中,关于这一点的一些提示也将受到赞赏.

Answer 1

最后,我确实为checkstyle实现了自定义检查.以下是其他人对此感兴趣的源代码:

import com.puppycrawl.tools.checkstyle.api.Check;
import com.puppycrawl.tools.checkstyle.api.DetailAST;
import com.puppycrawl.tools.checkstyle.api.TokenTypes;

public class IfBracesCheck extends Check {

    @Override
    public int[] getDefaultTokens() {
        return new int[] {
            TokenTypes.LITERAL_ELSE,
            TokenTypes.LITERAL_IF,
        };
    }

    @Override
    public void visitToken(DetailAST aAST) {
        final DetailAST slistAST = aAST.findFirstToken(TokenTypes.SLIST);

        if(aAST.getType() == TokenTypes.LITERAL_ELSE) {
            // If we have an else, it must have braces, except it is an "else if" (then the if must have braces).
            DetailAST ifToken = aAST.findFirstToken(TokenTypes.LITERAL_IF);

            if(ifToken == null) {
                // This is an simple else, it must have brace.
                if(slistAST == null) {
                    log(aAST.getLineNo(), "ifBracesElse", aAST.getText());
                }
            } else {
                // This is an "else if", the if must have braces.
                if(ifToken.findFirstToken(TokenTypes.SLIST) == null) {
                    log(aAST.getLineNo(), "ifBracesConditional", ifToken.getText(), aAST.getText() + " " + ifToken.getText());
                }
            }
        } else if(aAST.getType() == TokenTypes.LITERAL_IF) {
            // If the if uses braces, nothing as to be checked.
            if (slistAST != null) {
                return;
            }

            // We have an if, we need to check if it has no conditionnal structure as direct child.
            final int[] conditionals = {
                TokenTypes.LITERAL_DO,
                TokenTypes.LITERAL_ELSE,
                TokenTypes.LITERAL_FOR,
                TokenTypes.LITERAL_IF,
                TokenTypes.LITERAL_WHILE,
                TokenTypes.LITERAL_SWITCH,
            };

            for(int conditional : conditionals) {
                DetailAST conditionalAST = aAST.findFirstToken(conditional);

                if (conditionalAST != null) {
                    log(aAST.getLineNo(), "ifBracesConditional", aAST.getText(), conditionalAST.getText());

                    // Let's trigger this only once.
                    return;
                }
            }
        }
    }
}