我正在转换（从javascript）一个程序，该程序将采用可变长度的字符串（但始终低于100个字符），并以单个变量的形式返回字符串中包含的数据。这是我代码的第一部分，很明显，我对C和编程是个新手。这段代码是针对代码的第一部分，但学习如何正确编码这将使我知道如何编码其余部分
我需要：
- 要存储为"stringID"的前4位数字
- 要存储为"myindicator"的第5位数字
- 存储为"var1"的第6位至（指示符+6）位数字
示例输入：
'12345678901234567890123'
示例输出：
- 字符串ID=1234
- myindicator=5
- var1=67890123456
当我运行程序时，它返回"字符串ID:H>a"，然后程序崩溃。如有任何帮助，我们将不胜感激。不，这不是家庭作业。
int main()
{
char mystring[100];
char *stringID;
int nep;
int *myindicator;
char *var1;
nep = 0;
printf("Please enter your CODEn");
scanf("%s", &mystring);
stringID = (char *)malloc(4 * sizeof(char));
if(NULL != stringID)
{
strncpy(stringID, mystring, 4);
stringID[4] = ' ';
free(stringID);
nep = nep +4;
printf("stringID: %sn",myindicator);
}
if(NULL != myindicator)
{
strncpy(myindicator, (mystring+nep, 1);
nep++;
myindicator = *myindicator - '0';
printf("Indicator : %dn",myindicator);
}
var1 = (char *)malloc((nep + 6) * sizeof(char));
if(NULL != var1)
{
strncpy(var1, mystring+nep, (myindicator+nep+6));
var1[myindicator+nep+6] = ' ';
free(var1);
printf("Var 1: %s", var1);
nep = nep +myindicator+6;
}
getchar();
return 0;
}
我修复了一些问题，可以在注释中找到。但你需要查看C语言手册。。。！
int main()
{
char mystring[100];
char *stringID;
int nep;
// Changed to integer, not pointer to int.
int myindicator;
char *var1;
nep = 0;
printf("Please enter your CODEn");
/*
This scanf is a bad idea for the same reason for which, below, we take
care to allocate memory enough for whatever we have to do.
Should someone input 250 characters in a buffer of size 100, those 150
extra characters would wreak havoc and possibly endanger the system.
*/
// scanf("%s", &mystring);
fgets(mystring, sizeof(mystring)-1, stdin);
// fgets will read at most "sizeof(mystring)-1", that is, 99 bytes,
// from "stdin" (STanDard INput), the same as scanf. But it will halt
// when reaching the limit given. It's up to us to give a "real" limit
// (nothing stops you from saying 15000 -- even if the true value is 100).
// C strings are made of characters, terminated by a zero byte.
// So you need 5 here, to store 4 characters
stringID = (char *)malloc(5 * sizeof(char));
if (NULL == stringID)
{
// Serious out of memory error: no sense going on.
// fprintf(stderr, "Out of memoryn");
abort();
}
strncpy(stringID, mystring, 4);
stringID[4] = ' ';
printf("ID: %sn", stringID);
free(stringID);
nep = nep + 4;
printf("NEP: %dn", nep);
// Now we want to decode the fifth digit.
// I use '0' as character. So if the fifth digit is '0', '0'-'0' will give 0
// and if it is '9', '9'-'0' will give 9 (the number).
// The trick does not work with more than one digit, of course.
myindicator = mystring[nep] - '0';
// Had I wanted to read 3 digits, I would have had to copy them into a
// temporary buffer, add a zero in the fourth position, then run atol()
// on the resulting buffer: atol("12345 " /* A STRING */) = 12345 /* A NUMBER */;
printf("VLI : %dn", myindicator);
// Copy "myindicator" bytes, so alloc myindicator+1 chars
var1 = (char *)malloc((myindicator + 1) * sizeof(char));
// Check that var1 is not null and abort if it is
if (NULL == var1)
abort();
strncpy(var1, mystring + 6, myindicator);
var1[myindicator+1] = ' ';
// Moved this printf before the free. See why below.
printf("Prefix : %sn", var1);
// NEVER use a variable after you freed it!!!
// it might APPEAR to work, but will stab you in the back the first chance it gets.
// Good if paranoid habit: null a var as soon as you've freed it.
free(var1); var1 = NULL;
getchar();
return 0;
}
为什么要释放数组？在将它们从堆中释放出来之后，您正在引用它们。
您的代码将在以下位置出错：
-
您在哪里分配了
myindicator？strncpy（myindicator，（mystring+nep，1）；//它将在这里断开。
-
free（var1）；
printf（"前缀：%s"，var1）；//再次分段
-
再次点击此处
strncpy（var1，mystring+nep，（myindicator+nep+6））//你的mystring在哪里？
-
用CCD_ 2输入字符串是个可怕的想法。使用类似CCD_ 3的缓冲IO。
-
您正在将
mystring暴露在缓冲区溢出中。谁在阻止用户输入120字节的字符串？我可以用小心的跳转指令来编写我的恶意代码。