printf格式化等同于cpp的cout

Question

我目前是计算机科学专业的学生,​​今天我收到了一份非常普通的作业,应该用C++编写.直到今天我才学习C语言.这更像是盲目的任务.

在C中,我通常使用这个:

printf("\n\n\t%-30s %-7d liters\n\t%-30s %-7d liters\n\t%-30s %-7d km",
       "Current gasoline in reserve:",
       db.currentGas,
       "Total gasoline used:",
       db.usedGas,
       "Total travel distance:",
       db.usedGas);

由于赋值的条件是它应该用C++编写,这就是我尝试过的:

cout << setw(30) << "\n\n\tCurrent gasoline in reserve: "
     << setw(7) << db.currentGas << "litres"
     << setw(30) << "\n\tTotal gasoline used: "
     << setw(7) << db.usedGas << "litres"
     << setw(30) << "\n\tTotal travel distance: "
     << setw(7) << db.travelDistance << "km";

但看起来C %-30s和C++ 之间存在差异setw(30)？

Answer 1

确实存在差异,如下所示:

Georgioss-MacBook-Pro:~ gsamaras$ g++ -Wall main.cpp
Georgioss-MacBook-Pro:~ gsamaras$ ./a.out 


    Current gasoline in reserve:       6litres       
    Total gasoline used:       5litres     
    Total travel distance:       4kmGeorgioss-MacBook-Pro:~ gsamaras$ gcc -W
Georgioss-MacBook-Pro:~ gsamaras$ ./a.out 


    Current gasoline in reserve:   6       liters
    Total gasoline used:           5       liters
    Total travel distance:         4       kmGeorgioss-MacBook-Pro:~ gsamaras$ 

但问题是区别在哪里？

setw(30)相当于%30s,但是,你已经使用了-30s,左对齐输出!为了获得类似的行为,请使用std :: left,如下所示:

cout << "\n\n" << left << setw(30) << "\tCurrent gasoline in reserve: " << left << setw(7) << 6 << "litres\n" << left << setw(30) << "\tTotal gasoline used: " << left << setw(7) << 5 << "litres\n" << left << setw(30) << "\tTotal travel distance: " << left << setw(7) << 4 << "km";