strncpy | 易学教程

strncpy发生stack corruption detected(-fstack-protector)栈溢出

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代码: char line[MAX] = {0}; strncpy(line,pBeginObj,(ptemp - pBeginObj + 1)); log如下: 解释: char *strncpy(char *dest, const char *src, int n) 把src所指向的字符串中以src地址开始的前n个字节复制到dest所指的数组中 0x746d498e86 - 0x756d498cda = 1ac = 428 MAX = 256; 429>256,strncpy内存拷贝越界,导致栈溢出!!! 来源： CSDN 作者：羽落长安链接： https://blog.csdn.net/u012906122/article/details/103639970

What is the trick behind strcpy()/uninitialized char pointer this code?

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问题 #include <stdio.h> #include <string.h> #include <stdlib.h> void main () { char *imsi; unsigned int i; int val; char *dest; imsi = "405750111"; strncpy(dest,imsi,5); printf("%s",dest); /* i = 10; */ } In the above code, with the i = 10 assignment is commented as above, the code works fine without error. When assignment is included for compilation, the error (segmentation fault) occurs at strncpy(dest,imsi,5); . By avoiding optimization to variable i (i.e., volatile int i; ), the error is

strncpy doesn't always null-terminate

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问题 I am using the code below: char filename[ 255 ]; strncpy( filename, getenv( "HOME" ), 235 ); strncat( filename, "/.config/stationlist.xml", 255 ); Get this message: (warning) Dangerous usage of strncat - 3rd parameter is the maximum number of characters to append. (error) Dangerous usage of 'filename' (strncpy doesn't always null-terminate it). 回答1: I typically avoid using str*cpy() and str*cat() . You have to contend with boundary conditions, arcane API definitions, and unintended

strncpy or strlcpy in my case

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问题 what should I use when I want to copy src_str to dst_arr and why? char dst_arr[10]; char *src_str = "hello"; PS: my head is spinning faster than the disk of my computer after reading a lot of things on how good or bad is strncpy and strlcpy . Note: I know strlcpy is not available everywhere. That is not the concern here. 回答1: strncpy is never the right answer when your destination string is zero-terminated. strncpy is a function intended to be used with non-terminated fixed-width strings.

Why is vsnprintf Not Writing the Same Number of Characters as strncpy Would?

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问题 I've asked the same question about strncpy, but there the string ends up containing the whole input string. When passing a string to vsnprintf the last character always gets chopped off: https://rextester.com/UIQMX91570 For simplicity I've also included the live example link above inline in the code: void bar(const char* format, va_list vlist) { const auto buf_size = vsnprintf(nullptr, 0U, format, vlist); string buffer(buf_size, '\0'); vsnprintf(data(buffer), buf_size, format, vlist); cout <<

Is it Safe to strncpy Into a string That Doesn't Have Room for the Null Terminator?

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问题 Consider the following code: const char foo[] = "lorem ipsum"; // foo is an array of 12 characters const auto length = strlen(foo); // length is 11 string bar(length, '\0'); // bar was constructed with string(11, '\0') strncpy(data(bar), foo, length); cout << data(bar) << endl; My understanding is that string s are always allocated with a hidden null element. If this is the case then bar really allocates 12 characters, with the 12 th being a hidden '\0' and this is perfectly safe... If I'm

strcpy, strcpy_s, strncpy, strncpy_s整合

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strcpy, strcpy_s, strncpy, strncpy_s strcpy用法: strcpy_s用法： strncpy用法： strncpy_s用法： strcpy用法: 原型声明：char *strcpy(char* dest, const char *src); 头文件：#include <string.h> 和 #include <stdio.h> 功能：把含有’\0’结束符的字符串复制到另一个地址空间，返回值的类型为char*。说明：src和dest所指内存区域不可以重叠且dest必须有足够的空间来容纳src的字符串。 strcpy_s用法：该函数是VS2005之后的VS提供的，并非C标准函数原型：三个參数时：errno_t strcpy_s(char *strDestination, size_t numberOfElements, const char *strSource); 两个參数时：errno_t strcpy_s(char (&strDestination)[size], const char *strSource); // C++ only 功能：同strcpy()函数功能相同，不同之处在于参数中多了个size_t类型的参数，该参数为字符串dst的长度，当存在缓存区溢出的问题时（即src的长度大于dst的长度），strcpy

Strncpy in MIPS has a weird behavior

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问题 Here's my code for strncpy. In theory it should work, but when I run tests on it it gives out garbage. Arguments: $a0 = pointer to destination array $a1 = source string $a2 = number of characters to copy Returns: the destination array strncpy: beqz $a2, out lb $t0, 0($a1) #load byte beqz $t0 out subiu $a2, $a2, 1 sb $t0, 0($a0) addiu $a0, $a0, 1 addiu $a1, $a1, 1 j strncpy out: lb $0 0($a0) move $v0 $a0 jr $ra 回答1: coppy original address of destination array ($a0) and load it in "out:" in $a0

C: Create own version of strncpy

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问题 I am working on creating my own version of strncpy . My code seems to accept the input fine, but the program terminates after the input is entered. strncpy also seems to pad the copied function with nulls if it's shorter than the first - what is the point of that and how do I implement that in my code? #include <stdio.h> #include <stdlib.h> #define SIZE 50 #define STOP "quit" char *copywords(char *str1, char *str2, int n); int main(void) { char words[SIZE]; char newwords[SIZE]; int num; int i

format ’%s’ expects argument of type ’char *’

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问题 For exercising my programming skills in C I'm trying to write the strncpy function by myself. Doing that I kept hitting errors, solving most of them eventually I'm stuck with no further inspiration to go on. The error I receive is: ex2-1.c:29:3: warning: format ‘%s’ expects argument of type ‘char *’, but argument 2 has type ‘int’ [-Wformat=] printf("The copied string is: %s.\n", stringb); The thing is that it's a very common error and that it's also already described on SO, only I can't seem