strncpy

#include <stdio.h> #include <windows.h> #include <string.h> #include <assert.h> char * My_strncpy(char * dest, const char *src, size_t n) { assert(dest); assert(src); char *ret = dest; while (n--) { *dest++ = *src++; } return ret; } int main() { char str1[20] = "123456789"; char str2[20] = "abcde"; printf("调用My_strncpy之前: str1: %s \n", str1_1); My_strncpy(str1, str2, 5); //调用My_strncpy printf("调用My_strncpy之后: str1: %s \n", str1); system("pause"); return 0; } 文章来源: 模拟实现strncpy

I can see many sprintf 's used in my applications for copying a string. I have a character array: char myarray[10]; const char *str = "mystring"; Now if I want want to copy the string str into myarray , is is better to use: sprintf(myarray, "%s", str); or strncpy(myarray, str, 8); ? Neither should be used, at all. sprintf is dangerous, deprecated, and superseded by snprintf . The only way to use the old sprintf safely with string inputs is to either measure their length before calling sprintf , which is ugly and error-prone, or by adding a field precision specifier (e.g. %.8s or %.*s with an

how would you get the last word of a string, starting from the '\0' newline character to the rightmost space? For example, I could have something like this where str could be assigned a string: char str[80]; str = "my cat is yellow"; How would I get yellow? Something like this: char *p = strrchr(str, ' '); if (p && *(p + 1)) printf("%s\n", p + 1); mikithskegg I would use function strrchr() In case you don't want to use 'strrchr' function, Here is the solution. i = 0; char *last_word; while (str[i] != '\0') { if (str[i] <= 32 && str[i + 1] > 32) last_word = &str[i + 1]; i++; } i = 0; while

The function strncpy() doesn't always null terminate so I want to know what is the best alternative that always null terminates? I want a function that if: strlen(src) >= n /*n is the number of characters to be copied from source*/ there's no need to add further code like this: buf[sizeof(buf)-1] = 0; RoadRunner If the length of the string you desire to copy is unknown, you can use snprintf here. This function sends formatted output to str . It acts similarily to sprintf() , but instead does not write more bytes allocated by str . If the resulting string is longer than n-1 characters, then the

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. 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. More precisely, its purpose is to convert a zero-terminated string to a non-terminated fixed-width string (by

At the following regarding strncpy : http://www.cplusplus.com/reference/clibrary/cstring/strncpy/ , it mentions the following: No null-character is implicitly appended to the end of destination, so destination will only be null-terminated if the length of the C string in source is less than num. What is meant by this sentence? It means that, for example, if your source string is 20 characters plus a null terminator and your strncpy specifies less than 21 characters, the target string will not have a null appended to it. It's because of the way it works: strncpy guarantees that it will write

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 */ Others have pointed out your null-termination problems. You need to understand null-termination before you understand the difference between