[英]How to get the correct number of C++ command line arguments?
I start my program with 我从开始我的程序
int main (int argc, char *argv[]) {
printf("%d \n", argc);
then I compiled in Ubuntu using g++, and I ran the program using 然后我使用g ++在Ubuntu中编译,然后使用
./calc 2 3 4 + *
but the program outputs 17! 但是程序输出17! I did a printf on the arguments as well, they are:
我也对参数进行了printf,它们是:
arg 0: ./calc
arg 1: 2
arg 2: 3
arg 3: 4
arg 4: +
arg 5: 1.2.c
arg 6: 1.3.c
arg 7: 1.4.c
arg 8: 2.1.c
arg 9: 2.2.c
arg 10: 2.3.c
arg 11: 2.4.c
arg 12: 3.2.c
arg 13: 3.4.c
arg 14: 4.1.c
arg 15: a.out
arg 16: calc
but obviously that's not what I'm expecting. 但这显然不是我所期望的。 How can I correct this?
我该如何纠正?
The *
is being evaluated by your shell to mean all of the files in the current directory. Shell正在评估
*
表示当前目录中的所有文件。 You should escape the asterisk using \\*
. 您应该使用
\\*
转义星号。
The linux shell interpreted * as a listing of all the files in a directory. linux shell将*解释为目录中所有文件的列表。 Try escaping it with "\\" eg
尝试用“ \\”转义,例如
./calc 2 3 4 + \*
or 要么
./calc 2 3 4 + "*"
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