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[case]To prevent static resource search
王集鹄 edited this page Aug 4, 2015
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1 revision
通常静态资源的字符串是连续的,通过搜索关键字很容易定位并修改其内容。如果将连续的字符串分开赋值,那么定位就不能简单的搜索完成
#include<stdio.h>
int main(void)
{
/*<jdists export="#encode">
function (content) {
return content.replace(/char\s+(\w+)\[\s*(\d+)\s*\]\s*=\s*"(.*?)"/g,
function (all, name, len, value) {
var items = value.split('').concat([0]);
items = items.map(function (item, index) {
var value = item === 0 ? "'\\0'" : "'" + item + "'";
return '\t' + name + '[' + index + ']=' + value;
});
return 'char ' + name + '[' + len + '];\n' + items.join(';\n');
}
);
}
</jdists>*/
/*<jdists encoding="#encode">*/
char _link1[100] = "http://legend.baidu.com/";
char _link2[100] = "http://shushuo.baidu.com/";
/*</jdists>*/
printf("link1: %s\n link2", _link2, _link2);
return 0;
}
#include<stdio.h>
int main(void)
{
char _link1[100];
_link1[0]='h';
_link1[1]='t';
_link1[2]='t';
_link1[3]='p';
_link1[4]=':';
_link1[5]='/';
_link1[6]='/';
_link1[7]='l';
_link1[8]='e';
_link1[9]='g';
_link1[10]='e';
_link1[11]='n';
_link1[12]='d';
_link1[13]='.';
_link1[14]='b';
_link1[15]='a';
_link1[16]='i';
_link1[17]='d';
_link1[18]='u';
_link1[19]='.';
_link1[20]='c';
_link1[21]='o';
_link1[22]='m';
_link1[23]='/';
_link1[24]='\0';
char _link2[100];
_link2[0]='h';
_link2[1]='t';
_link2[2]='t';
_link2[3]='p';
_link2[4]=':';
_link2[5]='/';
_link2[6]='/';
_link2[7]='s';
_link2[8]='h';
_link2[9]='u';
_link2[10]='s';
_link2[11]='h';
_link2[12]='u';
_link2[13]='o';
_link2[14]='.';
_link2[15]='b';
_link2[16]='a';
_link2[17]='i';
_link2[18]='d';
_link2[19]='u';
_link2[20]='.';
_link2[21]='c';
_link2[22]='o';
_link2[23]='m';
_link2[24]='/';
_link2[25]='\0';
printf("link1: %s\n link2", _link2, _link2);
return 0;
}