[英]c++ Reading in integers from a .txt file to a stack
This is so stupid. 这太愚蠢了。 I've been stuck literally for an hour trying to read in a .txt file of numbers that are separated by a single whitespace. 我一直被困在一个小时内试图读取一个由单个空格分隔的.txt文件。 The while loops only gets executed once for some reason! while循环只能由于某种原因执行一次!
#include <iostream>
#include <string>
#include <fstream>
#include <stack>
using namespace std;
int main(int argc, char* argv[])
{
string line;
string str(argv[1]);
ifstream myfile((str).c_str());
int num;
stack<int> x;
while (myfile >> num);
{
x.push(num);
}
return(0);
}
Hmm, look at this line more closely: 嗯,更仔细地看一下这条线:
while (myfile >> num);
Eventually, you'll notice the semi-colon. 最终,你会注意到分号。 The compiler thinks this means you want a loop that does nothing (the semi-colon here indicates a single, empty statement). 编译器认为这意味着你想要一个什么都不做的循环(这里的分号表示一个空的语句)。 So, the loop reads in all the numbers, but does nothing with them. 因此,循环读取所有数字,但对它们不做任何操作。
The next section is interpreted separately as a statement in its own scope (denoted by the braces), to be executed after the loop: 下一节被单独解释为在其自己的范围内的语句(由大括号表示),在循环之后执行:
{
x.push(num);
}
All that does is push the last number read onto the stack, leading you to think the loop only executes once. 所有这一切都是将最后一个数字读入堆栈,导致您认为循环只执行一次。
Remove the ;
删除;
and you're OK! 你很好! Once bitten by this, you'll never forget ;-) 一旦被这个咬了,你永远不会忘记;-)
On an unrelated note, it's a bit silly to take argv[1]
(a C-style string), put it into a string
object, then use c_str()
to turn that back into a C-string for the ifstream constructor. 在一个不相关的说明中,采用argv[1]
(一个C风格的字符串),将它放入一个string
对象,然后使用c_str()
将其转换回ifstream构造函数的C字符串是有点傻。 Just use argv[1]
directly, since you're not doing anything else with it. 只需直接使用argv[1]
,因为你没有做任何其他事情。 Also, it would be a good idea to check argc
first and make sure that a filename was passed in. Finally, you should check that the file was successfully opened instead of assuming it -- at the very least make your assumption explicit with an assert(myfile.is_open());
此外,最好首先检查argc
并确保传入文件名。最后,您应检查文件是否已成功打开而不是假设 - 至少使用assert(myfile.is_open());
明确表示您的假设assert(myfile.is_open());
. 。 Oh, and you don't use the line
variable at all. 哦,你根本不使用line
变量。
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