printf 显示错误的数字

Question

char n[100];
int u[100];
int x[8];

scanf("%[^\n]",&n);
scanf("%d %d",&x,&u);


printf("%s\n",n); //print correct
printf("%d\n",x); //print not the exact number i input in scan
printf("%d\n",u);// print not the exact number i input in scan

if i input x and u for ex: 12345678 20 the output will print: random number

Answer 1

For %[^\\n] , scanf expects a pointer to a char (which is the first of enough char to receive the string), but &n is a pointer to an array of char . Change scanf("%[^\\n]",&n); to scanf("%[^\\n]", n); . Although n is an array of char , when used like this, it is automatically converted to a pointer to its first element, so a pointer to a char will be passed.

For %d , scanf expects a pointer to an int , but &x and &u are pointers to arrays of int . Change int u[100]; to int u; and int x[8]; to int x; . Then, in scanf("%d %d",&x,&u); , pointers to int will be passed.

In printf("%s\\n",n); , for %s , printf expects a pointer to the first of several char to print, and n is automatically converted to a pointer to its first element, so this is correct.

In printf("%d\\n",x); , for %d , printf expects an int . Previously, when x was an array, this was passing a pointer to the first int in the array. Then printf may have been printing the pointer as if it were an int , but a variety of things can go wrong with this. With the corrected definition of x shown above, this will pass the value of x instead of a pointer, so it will be correct.

Similarly, printf("%d\\n",u); will be correct once the definition of u is changed from an array to a single item.

Turn on warnings in your compiler and pay attention to them.