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Declare 'sum' so that it doesn't require type arguments. #17079

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Merged
DanielRosenwasser merged 2 commits into from
Jul 12, 2017
Merged

Declare 'sum' so that it doesn't require type arguments. #17079

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d25fd23
Declare 'sum' so that it doesn't require type arguments.
DanielRosenwasser Jul 10, 2017
325f4b8
Addressed feedback.
DanielRosenwasser Jul 11, 2017
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6 changes: 3 additions & 3 deletions src/compiler/checker.ts
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Original file line number Diff line number Diff line change
Expand Up @@ -83,9 +83,9 @@ namespace ts {
// extra cost of calling `getParseTreeNode` when calling these functions from inside the
// checker.
const checker: TypeChecker = {
getNodeCount: () => sum<"nodeCount">(host.getSourceFiles(), "nodeCount"),
getIdentifierCount: () => sum<"identifierCount">(host.getSourceFiles(), "identifierCount"),
getSymbolCount: () => sum<"symbolCount">(host.getSourceFiles(), "symbolCount") + symbolCount,
getNodeCount: () => sum(host.getSourceFiles(), "nodeCount"),
getIdentifierCount: () => sum(host.getSourceFiles(), "identifierCount"),
getSymbolCount: () => sum(host.getSourceFiles(), "symbolCount") + symbolCount,
getTypeCount: () => typeCount,
isUndefinedSymbol: symbol => symbol === undefinedSymbol,
isArgumentsSymbol: symbol => symbol === argumentsSymbol,
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5 changes: 3 additions & 2 deletions src/compiler/core.ts
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Expand Up @@ -700,10 +700,11 @@ namespace ts {
return result;
}

export function sum<K extends string>(array: { [x in K]: number }[], prop: K): number {
export function sum<T extends Record<K, number>, K extends string>(array: T[], prop: K): number {
let result = 0;
for (const v of array) {
result += v[prop];
// Note: we need the following type assertion because of GH #17069
result += v[prop] as number;
}
return result;
}
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