error checking help

[rcjay2]

Everyone,

I am having a problem with trying to get this to prompt the user for the correct input. I would like to user to be required to enter numbers only after it prompts him/her at "How much money would you like to convert today? (above 100): " My question is what is the easiest way to get this done?

Also, I am a newbie to this and this is my first programming class so please break things down so even I can understand them.

Thank you for your help.

[PHP]#include
int main (int argc, char** argv)

{

float name; //declares a variable "name"
float canada; //declares a variable "canada"
float sa; //declares a variable "sa"
float botswana; //declares a variable "botswana"
float china; //declares a variable "china"
float russia; //declares a variable "russia"
float amt; //declares a variable "amt"
int correctItemsRead;

printf("\n"); //Skips a line
printf("Currency Conversion\n"); //Displays Currency Conversion to screen
printf("\n"); //Skips a line
printf("I can convert US dollars to the following currencies: \n"); //Prints statement to the screen
printf("\n"); //Skips a line
printf("Canadian Dollars\nSouth African Rand\nBotswana Pula\nChina Renminbi\nRussian Rubles\n"); //Lists the different currencies that will be converted
printf("\n"); //Skips a line

do //This says to do what is in the braces at least once
{

printf("How much money would you like to convert today? (above 100): ");
correctItemsRead = scanf("%f", &amt);
}
while ((amt <= (float) 100) && (correctItemsRead == 1)); // this says that if amt <= 100 do what is in the braces again
//(float) tells the compiler to treat the value next to it
//like a float instead of a literal int


printf("I would be happy to help you with that, Thank you.\n"); //Prints statement to the screen

canada=1.324*amt; //Sets "canada" to 1.324 x "amt"
sa=6.865*amt; //Sets "sa" to 6.865 x "amt"
botswana=4.558*amt; //Sets "botswana" to 4.558 x "amt"
china=8.265*amt; //Sets "china" to 8.265 x "amt"
russia=29.234*amt; //Sets "russia" to 29.234 x "amt"

printf("\n"); //Skips a line

printf("In Canadian Dollars you would have = %.2f\n",canada); //Displays dollar equivilant to user input as Canadian Dollars
printf("In South African Rand you would have = %.2f\n",sa); //Displays dollar equivilant to user input as Canadian Dollars
printf("In Botswana Pula you would have = %.2f\n",botswana); //Displays dollar equivilant to user input as Canadian Dollars
printf("In China Renminbi you would have = %.2f\n",china); //Displays dollar equivilant to user input as Canadian Dollars
printf("In Russian Rubles you would have = %.2f\n",russia); //Displays dollar equivilant to user input as Canadian Dollars
printf("\n"); //Skips a line

printf("I would go for the Russian Rubles. At least you'll feel rich!\n"); //Prints statement to screen

printf("\n"); //Skips a line
printf("\n"); //Skips a line

printf("By the way, Could I get your name please, "); //Prints atatement to the screen and capture users name

scanf("%s",&name); //Captures user input
printf("\n"); //Skips a line
printf("\n"); //Skips a line
printf("\n"); //Skips a line
printf("\n"); //Skips a line
printf("\n"); //Skips a line
printf("You have a wonderful adventure,%s.\n",&name); //Prints users input to the screen


getchar();

return 0;
}[/PHP]

[infamous]

u are almost there. the scanf function returns teh number of correctly read variables. in your while loop, u are testing:
while ((amt <= (float) 100) && (correctItemsRead == 1));

but actually it should be:
while ((amt <= (float) 100) && (correctItemsRead != 1));

when it returns 1, u want to break out of the loop, not the other way around.

[gurusaranya]

hey you are really done pa...try it i think th latter is correct

[abhimanipal]

You have declared the name as a float variable. Change it to a char array

Also the correct way to accept a string via scanf statement is
scanf("%s",input);

PS: Using scanf to take user i/p is a bad idea. It usually leads to a wide varieties of problems. Use fgets for taking i/p from the user