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ISCH_PrimitivePin.isAsync() method

查询图元是否为异步图元

签名

public isAsync(): boolean;

返回值

boolean

是否为异步图元

示例

// 0. 确保当前文档是符号编辑器：优先复用测试符号，没有则新建后打开
const libUuid = await eda.lib_LibrariesList.getPersonalLibraryUuid();
const found = await eda.lib_Symbol.search('嘉立创示例_Pin测试符号', libUuid);
const symUuid = found[0] ? found[0].uuid : await eda.lib_Symbol.create(libUuid, '嘉立创示例_Pin测试符号');
await eda.lib_Symbol.openInEditor(symUuid, libUuid);

// 1. 在符号画布上创建一个测试引脚（默认异步模式）
const pin = await eda.sch_PrimitivePin.create(400, 300, '1', 'CLK', 0, 10, null, 'None', 'IN');
const asyncByDefault = pin.isAsync();

// 2. 转同步后再查，模式变为 false
const syncPin = pin.toSync();
const afterToSync = syncPin.isAsync();

// 3. 再转回异步，模式恢复 true
const asyncAgain = syncPin.toAsync();
const afterToAsync = asyncAgain.isAsync();

// 4. 清理测试引脚（查询类需要清理）
await eda.sch_PrimitivePin.delete([pin.getState_PrimitiveId()]);

console.log('asyncByDefault:', asyncByDefault);
console.log('afterToSync:', afterToSync);
console.log('afterToAsync:', afterToAsync);