我正在尝试在CMD开发人员提示符下运行应用程序,它应该具有使用命令行参数的输入文件和输出文件(我不理解,或者无法理解)。但是,当我这样做时,它正在运行并结束,但是没有任何内容被打印到输出文件中。
以下是我的代码,其中已删除了不相关的代码。我不知道我在代码或cmd行开发人员提示中做错了什么。
该应用程序被调用printlines.exe
,我的命令如下所示:
printlines.exe -i file.java -o output.txt
任何建议将不胜感激。
#include <stdio.h>
#include <string.h>
#include <stdbool.h>
#define LINE 1000
void countLines(int *emptyLines, int *totalLines, int *totalComments, char *fileName, FILE *readFile)
{
char line[LINE]; // set array to store input file data
//set variables
readFile = fopen(fileName, "r");
int lines = 0;
int comments = 0;
int empty = 0;
while (fgets(line, LINE, readFile) != NULL) //while loop that reads in input file line by line
{
/*counts lines in input file works when not using
command line arguments redacted to shorten */
}
fclose(readFile);
*emptyLines = empty;
*totalLines = lines;
*totalComments = comments;
}
int main(int argc, char *argv[])
{
FILE *inputFile;
FILE *outputFile;
char *outputName;
char *inputName;
inputName = argv[1];
outputName = argv[2];
inputFile = fopen(inputName, "r");
outputFile = fopen(outputName, "w");
int emptyLines, totalLines, totalComments;
countLines(&emptyLines, &totalLines, &totalComments, inputName, inputFile);
int character;
bool aComment = false;
while ((character = fgetc(inputFile)) != EOF)
{
/* code that writes info from input to output, works
when not using command line arguments redacted to shorten*/
}
//close files
fclose(inputFile);
fclose(outputFile);
printf("There are %d total lines, %d lines of code and %d comments in the file\n", totalLines, emptyLines, totalComments);
return 0;
}
命令行参数和打开文件的错误都不会执行错误检查。我敢打赌,如果您执行以下操作,您将迅速找出问题所在:
//....
if(argc < 5){ // check arguments
fprintf(stderr, "Too few arguments");
return EXIT_FAILURE;
}
inputFile = fopen(inputName, "r");
if(inputFile == NULL){ // chek if file was open
fprintf(stderr, "Failed to open input file\n");
return EXIT_FAILURE;
}
outputFile = fopen(outputName, "w");
if(outputFile == NULL){ // again
fprintf(stderr, "Failed to open output file\n");
return EXIT_FAILURE;
}
//...
请注意,您的comand行字符串具有5个参数,并且文件名位于索引2和4,因此您需要:
inputName = argv[2];
outputName = argv[4];
或只是删除-i
和,-o
因为它们似乎没有执行任何操作。
我还要注意,您可以fopen
在函数上或函数中直接使用命令行参数,不需要两个额外的指针:
inputFile = fopen(argv[2], "r");
outputFile = fopen(argv[4], "w");
//...
countLines(&emptyLines, &totalLines, &totalComments, argv[2], inputFile);
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