我在Qt中使用C ++。我有这个问题。我有import java.io.File;
import java.io.IOException;
import java.nio.charset.Charset;
import java.nio.file.Files;
import org.antlr.v4.runtime.ANTLRInputStream;
import org.antlr.v4.runtime.CommonTokenStream;
import org.antlr.v4.runtime.ParserRuleContext;
import org.antlr.v4.runtime.RuleContext;
import org.antlr.v4.runtime.tree.ParseTree;
public class ASTGenerator {
public static String readFile() throws IOException {
File file = new File("path/to/the/test/file.java");
byte[] encoded = Files.readAllBytes(file.toPath());
return new String(encoded, Charset.forName("UTF-8"));
}
public static void main(String args[]) throws IOException {
String inputString = readFile();
ANTLRInputStream input = new ANTLRInputStream(inputString);
Java8Lexer lexer = new Java8Lexer(input);
CommonTokenStream tokens = new CommonTokenStream(lexer);
Java8Parser parser = new Java8Parser(tokens);
ParserRuleContext ctx = parser.classDeclaration();
printAST(ctx, false, 0);
}
private static void printAST(RuleContext ctx, boolean verbose, int indentation) {
boolean toBeIgnored = !verbose && ctx.getChildCount() == 1 && ctx.getChild(0) instanceof ParserRuleContext;
if (!toBeIgnored) {
String ruleName = Java8Parser.ruleNames[ctx.getRuleIndex()];
for (int i = 0; i < indentation; i++) {
System.out.print(" ");
}
System.out.println(ruleName + " -> " + ctx.getText());
}
for (int i = 0; i < ctx.getChildCount(); i++) {
ParseTree element = ctx.getChild(i);
if (element instanceof RuleContext) {
printAST((RuleContext) element, verbose, indentation + (toBeIgnored ? 0 : 1));
}
}
}
}
（lineBox）和QLineBox（buttonGroup），其中包含QButtonGroup类型的按钮。当我在lineBox上按Enter键时，会为buttonGroup发出信号QPushButton。
lineBox位于buttonClicked。此QGraphicView与buttonGroup的布局相同。按Enter键时我不想发出clickButton。任何想法为什么会这样？
我的问题和这个问题有一些区别：How to make a QPushButton pressable for enter key?。它是在autoDefault属性的初始化中。