strncpy | 易学教程

strcpy与strncpy

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char aa[]="123456789123456789123456789"; char bb[4]={0}; 1、strcpy(bb,aa); bb的空间，不能存下aa的内容，导致踩到aa的内存。如何解决这个问题？ 2、使用strncpy，如下： strncpy(bb,aa,sizeof(bb)); 存在问题，bb的空间全部存储有效数据，没有预留\0的位置，strlen(bb)的长度是不确定的。如何解决这个问题？ 3、使用strncpy(bb,aa,sizeof(bb)-1); 4、注意：strncpy(bb,aa,n); 表面意思是拷贝n个字节。实际上，不是这样。表示最多拷贝n个字节。如下： char aa[]="123456789123456789123456789"; char bb[4]={0}; strncpy(bb,aa,3); //拷贝3个字节，bb为 1,2,3,\0 aa[1] = 0; strncpy(bb,aa,3); 按道理，应该拷贝3个字节，bb为1,\0,3,\0 实际上，不是这样，而只是拷贝了1,\0, bb为1,\0,\0,\0 也就是说，strncpy表示最多拷贝n个字节，来源的\0，会导致提前结束拷贝。转载于:https://www.cnblogs.com/nzbbody/p/4391644.html 来源： https://blog

Difference between strncpy and memcpy?

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问题 How can i access s[7] in s ? I didn't observe any difference between strncpy and memcpy . If I want to print the output s , along with s[7] (like qwertyA ), what are the changes I have to made in the following code: #include <stdio.h> #include <stdlib.h> int main() { char s[10] = "qwerty", str[10], str1[10]; s[7] = 'A'; printf("%s\n",s); strncpy(str,s,8); printf("%s\n",str); memcpy(str1,s,8); printf("%s\n",str1); return 0; } /* O/P qwerty qwerty qwerty */ 回答1: Others have pointed out your

strncpy leading to segmentation fault

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问题 I am just messing around with strncpy. My program looks like this typedef struct { char from_str[10]; }test; main () { test s1; memset(&s1,0,sizeof(test)); char src[10]="himansh"; char dest[10]; memset(dest,0,10); src[3]='\0'; printf("src is %s and strlen is %d \n", src,strlen(src)); fflush(stdout); strncpy(s1.from_str,src,100); printf("s1.from_str is %s , src is %s \n", s1.from_str,src); return 1; } Here before I do strncpy I have added a "\0" character in "src" string, length of "src"

使用strncpy注意事项

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1.strcpy是个不安全的函数，尽量使用strncpy替代 2.strncpy不拷贝'\0'，要注意详细解释见： https://blog.csdn.net/stpeace/article/details/22581763 来源： https://blog.csdn.net/wssjn1994/article/details/98957589

Why does strncpy not null terminate?

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strncpy() supposedly protects from buffer overflows. But if it prevents an overflow without null terminating, in all likelyhood a subsequent string operation is going to overflow. So to protect against this I find myself doing: strncpy( dest, src, LEN ); dest[LEN - 1] = '\0'; man strncpy gives: The strncpy() function is similar, except that not more than n bytes of src are copied. Thus, if there is no null byte among the first n bytes of src, the result will not be null-terminated. Without null terminating something seemingly innocent like: printf( "FOO: %s\n", dest ); ...could crash. Are

Why does strncpy not null terminate?

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问题 strncpy() supposedly protects from buffer overflows. But if it prevents an overflow without null terminating, in all likelyhood a subsequent string operation is going to overflow. So to protect against this I find myself doing: strncpy( dest, src, LEN ); dest[LEN - 1] = \'\\0\'; man strncpy gives: The strncpy() function is similar, except that not more than n bytes of src are copied. Thus, if there is no null byte among the first n bytes of src, the result will not be null-terminated. Without

Why is strncpy insecure?

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问题 I am looking to find out why strncpy is considered insecure. Does anybody have any sort of documentation on this or examples of an exploit using it? 回答1: Take a look at this site; it's a fairly detailed explanation. Basically, strncpy() doesn't require NUL termination, and is therefore susceptible to a variety of exploits. 回答2: The original problem is obviously that strcpy(3) was not a memory-safe operation, so an attacker could supply a string longer than the buffer which would overwrite

Why should you use strncpy instead of strcpy?

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问题 Edit: I\'ve added the source for the example. I came across this example: char source[MAX] = \"123456789\"; char source1[MAX] = \"123456789\"; char destination[MAX] = \"abcdefg\"; char destination1[MAX] = \"abcdefg\"; char *return_string; int index = 5; /* This is how strcpy works */ printf(\"destination is originally = \'%s\'\\n\", destination); return_string = strcpy(destination, source); printf(\"after strcpy, dest becomes \'%s\'\\n\\n\", destination); /* This is how strncpy works */

Why are strlcpy and strlcat considered insecure?

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问题 I understand that strlcpy and strlcat were designed as secure replacements for strncpy and strncat . However, some people are still of the opinion that they are insecure, and simply cause a different type of problem. Can someone give an example of how using strlcpy or strlcat (i.e. a function that always null terminates its strings) can lead to security problems? Ulrich Drepper and James Antill state this is true, but never provide examples or clarify this point. 回答1: Firstly, strlcpy has

Why is strncpy insecure?

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I am looking to find out why strncpy is considered insecure. Does anybody have any sort of documentation on this or examples of an exploit using it? Tim Take a look at this site ; it's a fairly detailed explanation. Basically, strncpy() doesn't require NUL termination, and is therefore susceptible to a variety of exploits. The original problem is obviously that strcpy(3) was not a memory-safe operation , so an attacker could supply a string longer than the buffer which would overwrite code on the stack, and if carefully arranged, could execute arbitrary code from the attacker. But strncpy(3)