strcpy

This question already has an answer here: warning: incompatible implicit declaration of built-in function ‘xyz’ 4 answers I just finnished my hangman game and as a last step I am doing some code cleanup and optimization, but I can't seem to understand why I receive the following two warnings: warning: incompatible implicit declaration of built-in function 'strlen' warning: incompatible implicit declaration of built-in function 'strcpy' The code in which they are used is here: for(i = 1; i <= random; i++) fgets(word, 100, f); fclose(f); for(i = 0; i < strlen(word); i++) if(word[i] == '\n') word

#include <stdio.h> #include <string.h> int main() { char src[]="123456"; strcpy(src, &src[1]); printf("Final copied string : %s\n", src); } When I use the Visual Studio 6 Compiler it gives me the expected answer " 23456 ". How come this program prints " 23556 " when compiled with gcc 4.7.2 ? strcpy(src, &src[1]); is undefined behavior: C11 §7.24.2.3 The strcpy function The strcpy function copies the string pointed to by s2 (including the terminating null character) into the array pointed to by s1 . If copying takes place between objects that overlap, the behavior is undefined. By the way,

This question already has an answer here: Is the function strcpy always dangerous? 9 answers I am a beginner, and I am learning how to copy a string in C now. Here is a problem I just met: Every time I try to use "strcpy" command to copy from string 1 to string 2, Visual Studio 2013 will give me an error/warning message saying that "strcpy" is unsafe and suggest me to use strcpy_s instead. Can you please explain why is strcpy unsafe to use? And what are the safer alternatives to strcpy? Here is my code: #include<stdio.h> #include<string.h> main() { char str1[] = "Copy a string."; char str2[15]

This question already has an answer here: Why do I get a segmentation fault when writing to a string initialized with “char *s” but not “char s[]”? 17 answers This works: int main() { char *t = "Hello"; t = "World"; printf("%s", t); } But this gives segmentation fault: int main() { char *t = "Hello"; strcpy(t, "World"); // the only difference printf("%s", t); } Why? Strings that you define explicitly - e.g. "Hello" - are typically placed in an area of read-only memory. These strings cannot be changed. In the first example, you are not changing the "Hello" string into the "World" string. You

#include<bits/stdc++.h> using namespace std; char *strcpy(char *a,const char *b) { if(a == NULL || b == NULL) exit(0); char *c = a; while((*a++ = *b++) != '\0'); return c; } int main() { char a[2] = "a"; char b[] = "bdfasdfdsfab"; strcpy(a,b); cout << a << endl; return 0; } 　　 来源： https://www.cnblogs.com/mch5201314/p/11684422.html