我的grunt文件中出现2个错误
line 61 col 25 This function's cyclomatic complexity is too high. (10)
line 101 col 22 This function's cyclomatic complexity is too high. (10)
在这种情况下如何降低Cyclomatic复杂度?我的功能并不复杂
第一个错误
remove: function(line, row, type) {
var spreadSelected = (row.spreadSelected && type === 'spread'),
totalSelected = (row.totalSelected && type === 'total'),
moneyLineSelected = (row.moneyLineSelected && type === 'moneyline'),
lineValue;
if (spreadSelected || totalSelected || moneyLineSelected) {
switch (type) {
case 'spread':
lineValue = row.spread.line;
break;
case 'total':
lineValue = row.total.line;
break;
case 'moneyline':
lineValue = row.moneyLineId;
break;
default:
break;
}
AuthFactory.getCustomer().then(function(credentials) {
betSlipSelectionRequest('/betSlip/removeSelection', {
customerId: credentials.customer,
game: row.game,
pair: row.pair,
line: lineValue
});
});
if (spreadSelected) {
row.spreadSelected = false;
}
if (totalSelected) {
row.totalSelected = false;
}
if (moneyLineSelected) {
row.moneyLineSelected = false;
}
}
}...
然后是第二个错误函数
add: function(line, row, type) {
var spreadSelected = (row.spreadSelected && type === 'spread'),
totalSelected = (row.totalSelected && type === 'total'),
moneyLineSelected = (row.moneyLineSelected && type === 'moneyline'),
lineValue;
if (!(spreadSelected || totalSelected || moneyLineSelected)) {
switch (type) {
case 'spread':
lineValue = row.spread.line;
break;
case 'total':
lineValue = row.total.line;
break;
case 'moneyline':
lineValue = row.moneyLineId;
break;
default:
break;
}
AuthFactory.getCustomer().then(function(credentials) {
betSlipSelectionRequest('/betSlip/addSelection', {
customerId: credentials.customer,
game: row.game,
pair: row.pair,
line: lineValue
});
});
switch (type) {
case 'spread':
row.spreadSelected = true;
break;
case 'total':
row.totalSelected = true;
break;
case 'moneyline':
row.moneyLineSelected = true;
break;
}
}
}
这里奇怪的是:这个错误只出现在我身上,我的同事打开相同的文件并运行grunt,终端没有错误。
答案 0 :(得分:3)
降低函数的圈复杂度的方法是将其拆分为几个较小的函数,并将这种复杂性分配到易于理解的块中。例如,您可以提取switch-case语句,结果如下:
remove: function(line, row, type) {
var spreadSelected = (row.spreadSelected && type === 'spread'),
totalSelected = (row.totalSelected && type === 'total'),
moneyLineSelected = (row.moneyLineSelected && type === 'moneyline'),
lineValue;
if (!(spreadSelected || totalSelected || moneyLineSelected)) {
lineValue = getLineValue(row, type);
}
// ... and so on, in reasonable chunks.
}
function getLineValue(row, type) {
var lineValue;
switch (type) {
case 'spread':
lineValue = row.spread.line;
break;
case 'total':
lineValue = row.total.line;
break;
case 'moneyline':
lineValue = row.moneyLineId;
break;
default:
break;
return lineValue;
}
然后,我们发现您可以在第二个块中重用getLineValue
函数:
add: function(line, row, type) {
var spreadSelected = (row.spreadSelected && type === 'spread'),
totalSelected = (row.totalSelected && type === 'total'),
moneyLineSelected = (row.moneyLineSelected && type === 'moneyline'),
lineValue;
if (!(spreadSelected || totalSelected || moneyLineSelected)) {
lineValue = getLineValue(row, type);
}
// ... and so on
}
因此,通过这一改变,您已将复杂性分配给另一个功能,并通过消除重复完全消除了一些复杂性。