append can be combined into a single call CRT-P000169 }
70
71 row := ui.TableRow{}
72 row = append(row, ui.TableCell{Contents: p.Name})73
74 row = append(row, pausedColumn)
75 85 }
86
87 row := ui.TableRow{}
88 row = append(row, ui.TableCell{Contents: j.Name}) 89 row = append(row, pausedColumn)
90 row = append(row, pausedByColumn)
91 row = append(row, pausedAtColumn) 95 row = append(row, ui.TableCell{Contents: p.TeamName})
96 }
97
98 row = append(row, pausedColumn) 99 row = append(row, pausedByColumn)
100 row = append(row, pausedAtColumn)
101 table.Data = append(table.Data, row) 88 }
89
90 row := ui.TableRow{}
91 row = append(row, ui.TableCell{Contents: strconv.Itoa(p.ID)}) 92 row = append(row, ui.TableCell{Contents: p.Ref().String()})
93
94 if command.All { 84 }
85
86 row := ui.TableRow{}
87 row = append(row, ui.TableCell{Contents: strconv.Itoa(p.ID)}) 88 row = append(row, ui.TableCell{Contents: p.Ref().String()})
89 if command.All {
90 row = append(row, ui.TableCell{Contents: p.TeamName})Multiple calls for append could be combined into a single call for append.
The append built-in function appends elements to the end of a slice. If it has
sufficient capacity, the destination is resliced to accommodate the new
elements, but if capacity is not enough, then append will allocate a new
underlying array and return the updated slice. Therefore, it is necessary to
store the result of append, often in the variable holding the slice itself.
Appending calls in a single call of append allocates memory just once to
accommodate all the elements to be appended. Whereas multiple calls to append
introduce many overheads, most notably being the possibility of more calls for
memory allocation because the total number of elements to be appended over
multiple calls of append is unknown beforehand, resulting in inaccurate
preallocation.
Read more about a
similar implementation for append.
Hence, a single call to append is recommended.
Bad practice
xs = append(xs, 1)
xs = append(xs, 2)
xs = append(xs, 3)
xs = append(xs, 4)
xs = append(xs, 5)
Recommended
xs = append(xs, 1, 2, 3, 4, 5)