In C, how do I check if a String contains 2 digits, 1 letter and 4 digits?
In the method I've tried this:
int 1, 2, 4, 5, 6, 7;
char 3;
char display[10];
scanf("%d%d%c%d%d%d%d", &1, &2, &3, 开发者_JAVA技巧&4, &5, &6, &7);
display = {1, 2, 3, 4, 5, 6, 7};
But I get errors everywhere and it doesn't work.
Presumably in that order?
Walk the string a char at a time and use isdigit() isalpha() to check each one.
Or just do:
char test[] = "12B3456";
if ( (strlen(test)>6) &&
isdigit(test[0]) &&
isdigit(test[1]) &&
isalpha(test[2]) &&
isdigit(test[3]) &&
isdigit(test[4]) &&
isdigit(test[5]) &&
isdigit(test[6]) )
{
// valid
}
First of all, in C, variable names can't start with a number, or be a number for that matter. So the declaration of int 1,2,3,4,5,6,7 will not compile, as well as char 3;
Here's a sample of how you could do it assuming the input is a null terminated string:
int matches(char *input){
int i;
/* This array contains 1 in places where a digit is expected */
char expected_digits[] = {1,1,0,1,1,1,1};
for(i = 0 ; input[i] != 0 && i < 7; i++){
if(expected_digits[i] == 1){
if(!isdigit(input[i])){
return 0;
}
}
else{
if(!isalpha(input[i]))
{
return 0;
}
}
}
if(i == 7) {
/* We reached the end of the input string and all its places matched */
return 1;
}
else{
return 0;
}
}
Not the best piece of code, but should do the trick. And it should compile with a C compiler.
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