@eagleoutice/flowr - v2.15.8
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    Variable DfgVertexConst

    DfgVertex: {
        hasOrigin(
            this: void,
            vertex: DataflowGraphVertexBase | undefined,
            origin: BuiltInProcName,
        ): boolean;
        is<T extends VertexType>(
            this: void,
            vertex: DataflowGraphVertexBase | undefined,
            tag: T,
        ): vertex is VertexByType[T];
        isFunctionCall(
            this: void,
            vertex?: DataflowGraphVertexBase,
        ): vertex is DataflowGraphVertexFunctionCall;
        isFunctionDefinition(
            this: void,
            vertex?: DataflowGraphVertexBase,
        ): vertex is DataflowGraphVertexFunctionDefinition;
        isUse(
            this: void,
            vertex?: DataflowGraphVertexBase,
        ): vertex is DataflowGraphVertexUse;
        isValue(
            this: void,
            vertex?: DataflowGraphVertexBase,
        ): vertex is DataflowGraphVertexValue;
        isVariableDefinition(
            this: void,
            vertex?: DataflowGraphVertexBase,
        ): vertex is DataflowGraphVertexVariableDefinition;
        name: "DfgVertex";
    } = ...

    The one helper to ask a vertex what it is.

    Every check tolerates an absent vertex, which is what DataflowGraph#getVertex|getVertex() hands back for an id the graph does not know.

    Type Declaration

    const vertex = graph.getVertex(id);
    DfgVertex.isFunctionCall(vertex) && vertex.name;     // narrows, so `name` is there
    DfgVertex.is(vertex, VertexType.Use);                // the same check from a tag
    DfgVertex.hasOrigin(vertex, 'builtin:eval');         // the call is an eval

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