AuthorizationExecuteWithPrivileges（）作为root错误
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我是可可的初学者。
我只想在我的Cocoa App中启动Apache和其他进程。
这是我的代码：
OSStatus myStatus;
AuthorizationFlags myFlags = kAuthorizationFlagDefaults;
AuthorizationRef myAuthorizationRef;
FILE *pipe = NULL;
myStatus = AuthorizationCreate(NULL, kAuthorizationEmptyEnvironment, myFlags, &myAuthorizationRef);
AuthorizationItem myItems = {kAuthorizationRightExecute, 0, NULL, 0};
AuthorizationRights myRights = {1, &myItems};
myFlags = kAuthorizationFlagDefaults |
kAuthorizationFlagInteractionAllowed |
kAuthorizationFlagPreAuthorize |
kAuthorizationFlagExtendRights;
myStatus = AuthorizationCopyPrivilegedReference (&myAuthorizationRef,kAuthorizationFlagDefaults);
myStatus = AuthorizationCopyRights (myAuthorizationRef,&myRights, NULL, myFlags, NULL );
char *tool = \"/usr/sbin/apachectl\";
char *args[] = { \"start\",NULL} ;
myStatus = AuthorizationExecuteWithPrivileges(myAuthorizationRef, tool, kAuthorizationFlagDefaults, args, &pipe);
char c[100];
int n=fread(c,1,100,pipe);
c[n] = \'\\0\';
NSLog(@\"%s\\n\",c);
没有找到相关结果
已邀请:
3 个回复
傻寺俊擒
草票
#include <stdio.h> int main(int argc, char** argv) { if (argc < 2) { printf(\"not enough arguments\\n\"); return -1; } if (0 != setuid(0)) { printf(\"setuid failed.\\n\"); return -3; } int i; char** argvz = (char**)malloc(sizeof(char*) * (argc - 1)); for (i = 1; i < argc; i++) { argvz[i - 1] = argv[i]; } execv(argv[1], argvz); printf(\"execv returned?\\n\"); return -2; }它将setuid作为root，然后使用给定的args运行有问题的程序。 请注意，您必须从AuthorizationExecuteWithPrivileges调用setuid，因为只有AuthorizationExecuteWithPrivileges创建的pid才具有升级权限（setuid将执行并用您自己的进程替换）。犁攀富