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[英]How to integrate an open source C program instead of calling its executable through a system call?
[英]system call open() creating executable
if(cmds.at(i)==">")
{
//convert strings to char*s
char* conversion = new char[cmds.at(i-1).size()+1];
copy(cmds.at(i-1).begin(),cmds.at(i-1).end(),conversion);
conversion[cmds.at(i-1).size()] = '\0';
const char * out_file_cstring = cmds.at(i+1).c_str();
//count and agregate arguments
int size = count_arguments(conversion);
size++;
char** args= new char*[size];//dont forget to delete
args[0] = strtok(conversion," \n");
for(int j = 1; j<size; j++){args[j] = strtok(NULL, " \n");}
args[size-1]= NULL;
//forking and redirection
int out_file = open(out_file_cstring,O_CREAT|O_WRONLY|O_TRUNC);
pid_t pid = fork();
if(!pid)
{
dup2(out_file,STDOUT_FILENO);
close(out_file);
int r = execvp(args[0],args);
if(r<0)
{
cerr<<"ERROR : exec failed"<<args[0]<<endl;
return false;
}
}
因此,我的代码会正确创建并写入out_file。 但是,由于某种原因,该文件是可执行文件。 我认为故障出在我的open()调用中,但我似乎找不到原因。
man 2 open
解释了原因:
int open(const char *pathname, int flags);
int open(const char *pathname, int flags, mode_t mode);
O_CREAT:
If the file does not exist it will be created. [...]
The permissions of the created file are (mode & ~umask).
因此,如果您想要一个不可执行的文件,则可以使用:
open(out_file_cstring,O_CREAT|O_WRONLY|O_TRUNC, 0666);
0666
将建议所有读/写(等同于常量标志S_IRUSR | S_IWUSR | S_IRGRP | S_IWGRP | S_IROTH | S_IWOTH
),并且用户的umask会将最终权限进一步限制为用户选择的默认值。
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