How to copy char *str to char c[] in C?

问题:

Trying to copy a char *str to char c[] but getting segmentation fault or invalid initializer error.

Why is this code is giving me a seg fault?

char *token = "some random string"; char c[80];   strcpy( c, token); strncpy(c, token, sizeof c - 1);  c[79] = '\0'; char *broken = strtok(c, "#"); 

回答1:

use strncpy() rather than strcpy()

/* code not tested */ #include   int main(void) {   char *src = "gkjsdh fkdshfkjsdhfksdjghf ewi7tr weigrfdhf gsdjfsd jfgsdjf gsdjfgwe";   char dst[10]; /* not enough for all of src */    strcpy(dst, src); /* BANG!!! */   strncpy(dst, src, sizeof dst - 1); /* OK ... but `dst` needs to be NUL terminated */       dst[9] = '\0';   return 0; } 

回答2:

char *str = "Hello"; char c[6]; strcpy( c, str ); 

回答3:

use strncpy to be sure to not copy more charachters than the char[] can contains

char *s = "abcdef"; char c[6];  strncpy(c, s, sizeof(c)-1); // strncpy is not adding a \0 at the end of the string after copying it so you need to add it by yourself c[sizeof(c)-1] = '\0'; 

Edit: Code added to question

Viewing your code the segmentation fault could be by the line

strcpy(c, token) 

The problem is if token length is bigger than c length then memory is filled out of the c var and that cause troubles.

回答4:

char c[] must have some size;

for example

char c[]= "example init string"; 

// that set up table c to c[19]; You can allocate it directly at the begginign of Your program;

char c[19] = {0}; // null filled table 

char c[i] is the pointer so You don't need to copy anything; char c[19] ; c = "example init string"; // now &c[0] points the same address;

Copy can be done wiht

 strcpy(dst, src); 

but MS force You to use secure function:

strcpy_s(dst,buffsize,src); 

回答5:

It's been a while since i coded in c/c++, but c[80] is probably allocated on the stack. If you use char *c and strdup or similiar you get it allocated on the heap where strtok can access it.

Try something like this.

char *token = "some random string"; char *c; c = strdup(token); char *broken = strtok(c, "#"); free(c); 

回答6:

Edited: Thanks for adding the code.

Perhaps the segfault occurs here:

strncpy(c, token, sizeof c - 1);  

sizeof has a the same precedence as '-' from right to left, so it is probably processed as :

strncpy(c, token, sizeof( c - 1 ) );  

instead of

strncpy(c, token, sizeof(c) - 1);  

which is probably what you wanted

(reference: http://en.wikipedia.org/wiki/Operators_in_C_and_C%2B%2B#Operator_precedence)

文章来源: How to copy char *str to char c[] in C?