C连接字符串不起作用

Question

int main(int argc, char** argv) {
    char data[1024];
    data[0] = '\0';
    for(int i = 1; i < argc; i++){
        strcpy(data+strlen(data), (argv[i] + 1));
    }
    strcpy(data+strlen(data), data+strlen(data)/2);
    printf(data);

    return 0;
}

As you can see this is my code so far. What I'm trying to do is: Remove first letter from every argument, concat them into data and after the loop take half of the resulting string and concat it again, then print it. Example:

Calling the program with the arguments hello , world and yes should print: elloorldesrldes

it works until strcpy(data+strlen(data), data+strlen(data)/2); . Here I try to take half of the string ( data ) and concat it to the end of the same string. When I leave that part out I get the result elloorldes but when I put it in, instead of giving me the expected results I get the error RUN FAILED (exit value -1.073.741.819, total time: 4s) , however I'm not sure why that's the case.

Answer 1

You cannot do this

strcpy(data+strlen(data), data+strlen(data)/2);

because strcpy cannot handle cases when memory overlaps.

man strcpy

 char *strcpy(char *dest, const char *src); 

DESCRIPTION

The strcpy() function copies the string pointed to by src , including the terminating null byte ( '\\0' ), to the buffer pointed to by dest . The strings may not overlap , and the destination string dest must be large enough to receive the copy.

You need to use memmove for this, which handles memory overlap:

size_t oldsize = strlen(data);
size_t size = oldsize/2;

memmove(data+oldsize, data+size, size);
data[oldsize + size] = 0;

Also don't do printf(data) with content provided by the user. Let's say the passed arguments are hello , world%d , then data will contain %d and printf would yield undefined behaviour, because there are arguments missing.

You should do this:

printf("%s\n", data);

or

puts(data);