How can I append one single char at the end of a new filename?-CodePudding

int main(int argc, char *argv[]) {

    int fd_src = open(argv[1], O_RDWR);

    ssize_t size = lseek(fd_src, 0, SEEK_END);

    char *buf[size];

    char *prefix = strcat(argv[1], "_");

    for(int i = 0; i < size-1; i  ) {

        lseek(fd_src, i, SEEK_SET);

        read(fd_src, buf, 1);

        char *postfix = &buf[0];

        char *filename = strcat(prefix, postfix);

        int fd_dest = open(filename, O_RDWR | O_CREAT | O_EXCL, 0644);

        write(fd_dest, buf, 1);

        close(fd_dest);

    }

    close(fd_src);

    return 0;

}

The input-file "filename" only contains the string "abc"

After I run the program, the output looks like that:

filename_a // file contains a filename_ab // file contains b filename_abc // file contains c

But I would like it to look like that:

filename_a // file contains a filename_b // file contains b filename_c // file contains c

I have already tried to declare the postfix like that:

char *postfix = &buf[i]

But that didn't work either. Any suggestions? :)

CodePudding user response：

char *prefix = strcat(argv[1], "_");

it is wrong as you attach one character to the string kept in the argv[1] and argv[1] is very unlikely to be big enough to accommodate it. It is very likely undefined behaviour.

From your code, I understand that you want to append the file name the first character of the file content.

char filename[strlen(argv[1]   3];
/* ... */
sprintf(filename, "%s_%c", argv[1], buf[0]);  // you need to change the type of buf

char *buf[size];

it is an array of pointers not characters. It will work as sizeof(char *) is always > sizeof(char), but you need an array of characters.

char buf[size];