strcpy

1. string转const char* string s = "abc" ; const char * c_s = s . c_str (); 2. const char*转string 直接赋值即可 const char * c_s = "abc" ; string s ( c_s ); 3. string转char* string s = "abc" ; char * c ; const int len = s . length (); c = new char [ len + 1 ]; strcpy ( c , s . c_str ()); 4. char*转string char * c = "abc" ; string s ( c ); 5. const char*转char* const char * cpc = "abc" ; char * pc = new char [ 100 ]; //足够长 strcpy ( pc , c） 语法: const char *c_str(); c_str()函数返回一个指向正规C字符串的指针, 内容与本string串相同. 这是为了与c语言兼容，在c语言中没有string类型，故必须通过string类对象的成员函数c_str()把string 对象转换成c中的字符串样式。 注意：一定要使用strcpy()函数 等来操作方法c

How do you concatenate or copy char* together? char* totalLine; const char* line1 = "hello"; const char* line2 = "world"; strcpy(totalLine,line1); strcat(totalLine,line2); This code produces an error! segmentation fault I would guess that i would need to allocate memory to totalLine? Another question is that does the following copy memory or copy data? char* totalLine; const char* line1 = "hello"; totalLine = line1; Thanks in advance! :) I would guess that i would need to allocate memory to totalLine? Yes, you guessed correctly. totalLine is an uninitialized pointer, so those strcpy calls are

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 to apply the tips other people have already pointed out. I get that I'm using a wrong type when

I need to store the input from a user into an array of strings. #include <stdlib.h> #include <stdio.h> #include <string.h> char *history[10] = {0}; int main (void) { char input[256]; input = "input"; strcpy(history[0], input); return (EXIT_SUCCESS); } Running it on the terminal I get a Segmentation Fault and in NetBeans I get main.c:11: error: incompatible types in assignment. I also tried to shift all the history to store the newest input into the first position (history[0]). history[9] = history[8]; history[8] = history[7]; history[7] = history[6]; history[6] = history[5]; history[5] =

In C i used strcpy to make a deep copy of a string, but is it still 'fine' to use strcpy in C++ or are there better alternatives which i should use instead ? In C++ the easiest way is usually to use the std::string class instead of char*. #include <string> ... std::string a = "Hello."; std::string b; b = a; The line "b = a;" does the same thing you would usually do with strcpy. I put this in the comment above, but just to make the code readable: std::string a = "Hello."; std::string b; b = a.c_str(); // makes an actual copy of the string b = a; // makes a copy of the pointer and increments the

Following is the most popular implementation of strcpy in traditional systems. Why dest and src are not checked for NULL in the start? I heard once that in old days the memory was limited so short code was always preferred. Will you implement strcpy and other similar functions with NULL pointer checks at the start now days? Why not? char *strcpy(char *dest, const char *src) { char *save = dest; while(*dest++ = *src++); return save; } NULL is a bad pointer, but so is (char*)0x1 . Should it also check for that? In my opinion (I don't know the definitive reason why), sanity checks in such a low

I'm a beginner to C but I have this code running on xcode through gcc on terminal: #include <stdio.h> #include <string.h> int main(){ char name[12] = "Roman Mirov"; printf("My name is %s\n", name); name[8] = 'k'; printf("My name is %s\n", name); char greeting[] = "hello"; printf("%s %s\n", greeting, name); strcpy(greeting, "greetings, "); printf("%s%s\n", greeting, name); return 0; } And it outputs this: My name is Roman Mirov My name is Roman Mikov hello Roman Mikov Abort trap: 6 My question exactly is, why it generates error instead of showing the last line as output "greetings, Roman Mikov"