C-通过FGETS()接受用户的多行输入

Question

我正在尝试使用fgets从用户那里接受多行，但是当我离开时，我的 Segmentation fault (core dumped)。我可以在循环中毫无疑问的情况下 printf address和 part_of_address变量，而在循环中，它可以按预期工作。一旦摆脱了循环，它就会着火。

// Define a char array called 'name' accepting up to 25 characters.
char name[25];
// Define a char array called 'part_of_address' accepting up to 80 characters.
char part_of_address[80];
// Define a char array called 'address' accepting up to 80 characters.
char address[80];
// Clean the buffer, just to be safe...
int c;
while ((c = getchar()) != 'n' && c != EOF) {};
// Ask for the user to enter a name for the record using fgets and stdin, store
// the result on the 'name' char array.
printf("nEnter the name of the user (RETURN when done):");
fgets(name, 25, stdin);
// Ask for the user to enter multiple lines for the address of the record, capture
// each line using fgets to 'part_of_address'
printf("nEnter the address of the user (DOUBLE-RETURN when done):");
while (1)
{
    fgets(part_of_address, 80, stdin);
    // If the user hit RETURN on a new line, stop capturing.
    if (strlen(part_of_address) == 1)
    {
        // User hit RETURN
        break;
    }
    // Concatinate the line 'part_of_address' to the multi line 'address'
    strcat(address, part_of_address);
}
printf("This doesn't print...");

Answer 1

正如迈克尔·沃尔茨（Michael Walz）注释时所指出的，即使在没有初始化的address的情况下，您也会使用strcat(address, part_of_address);。由于它是一个自动阵列，因此包含不确定的值，您正在调用未定义的行为。即使是第一个strcat也可能会在address数组之后覆盖内存。

只需使用char address[80] = "";或char address[80] = {''};