如何在C中的char数组中存储2位整数值和浮点值

Question

Following is my C code to print values of char but I am getting unexpected results. The code is

#include<stdio.h>

main()
{

  char sendBuffer[1000];
  float count;
  int i;
  for(count=1.5;count<=2.5;)
  {
   for(i=0;i<=15;)
   {
     sendBuffer[0]=count+48;
     sendBuffer[1]='a';
     sendBuffer[2]='b';
     sendBuffer[3]='c';
     sendBuffer[4]=i+48;
     sendBuffer[5]='\0';
     printf("%s\n",sendBuffer);
     i=i+5;
    }
   count=count+0.5;
  }
}

The results I am getting are:

1abc0
1abc5
1abc:
1abc?
2abc0
2abc5
2abc:
2abc?
2abc0
2abc5
2abc:
2abc?

whereas, I was expecting something like

1.5abc0
1.5abc5
1.5abc10
1.5abc15 

and so on. Can anybody tell me how to store integer and float values in char array in C?

Answer 1

I see the answers you're getting answer the problem, but don't give an explanation, and from the looks of it, you could use a good explanation, so here's what's happening:

Types in C

You need to understand that different types ( int , char , float , double ) take up different sizes in memory. In C, a char is always considered to take up a single byte, and is in the range of -127 to 128. ( unsigned char goes from 0 to 255.) int s take up 4 bytes, and their range is from -2M to 2M - 1. float and double take 4 and 8 bytes respectively, and have a huge range, but limited precision. (For more information, just search for "c type ranges", and you'll find lots of information, including these links on Stack Overflow, which also give you good explanations: Definition of range of a data type , Guarantateed minimum size/range of C data types ).

Also, you know enough ascii to know to add 48 to a digit to get the char version, but make sure you have an ascii table nearby. Conveniently, you can find one at http://www.asciitable.com .

You also have to understand that the compiler will do its best to convert from one to another, though in some cases, such as casting a float to an int , if you don't do it explicitly, your compiler should at least issue a warning.

Strings don't exist as a type in C. Instead, you print into an array. Other answers tell you how to do that. What their code is doing is printing the values you want into a character array, which is all you need.

What your code is doing

Now, let's see what your current code is doing:

First of all, a quick sidebar: when you're doing a for loop, it's best to put the increment in the loop itself, so:

for(count=1.5;count<=2.5;count=count+0.5)
{
  for(i=0;i<=15;i=i+5)

would be more standardized.

Now, your first line is sendBuffer[0]=count+48; . Here's what happens:

1. You're taking count, which starts at 1.5, and adding 48, so the first number is 49.5.
2. The compiler will automatically change a float to an int , though it should give you a warning about it. It does so by truncating the value, changing 49.5 to 45.
3. It then changes (int)49 to (char)49 which is the same value, but takes up one byte instead of 4. Ascii 49 is '1', so this happens to be the number you wanted.
4. The compiler does not do anything with the rest of the float. It never sets sendBuffer[1] or [2] for this value.

Your next 3 lines hard code the second, third and fourth characters of the string to abc . Even if your first line had set the full float value into sendBuffer , this would overwrite it.

Your final line takes i, which will be 0, 5, 10 and 15, and adds it to 48. This is just like the first step, without the truncation from float . So first it looks up ascii 48, which is '0', then ascii 53, which is '5'. When i is 10, it now takes the int 58 and converts it to a character (again, changing it from 4 bytes to 1 byte). Printing ascii 58 results in ':', and printing ascii 63 results in '?'. And that's how your output ended up the way it did.

Possible solutions

Finally, a few words on the solution you were given:

Since all you're doing is outputting the result, you don't need sendBuffer nor snprintf . Simply using

printf("%.1fabc%d\n", count, i);

would have sufficed.

It's instructive to think about what the printf family of functions does, though. Though it's less efficient, let's use a few snprintf s to see what it's doing:

int index = 0;
index = snprintf(&sendbuf[index], sizeof(sendbuf) - index, "%.1f", count);
index += snprintf(&sendbuf[index], sizeof(sendbuf) - index, "abc");
index += snprintf(&sendbuf[index], sizeof(sendbuf) - index, "%d", i);
sendbuf[index++] = '\n';
sendbuf[index] = '\0';

What this does is what *printf does when you pass it multiple arguments: keep track of how much you've written and use that to tell where you're writing the next parts.

If you don't know it, the a += b syntax is shorthand for a = a + b . a++ means "give me the value of a, then increment it."

So the second line of my code will print out the float value into sendbuf, starting at the 0 position. *printf always returns the number of characters printed, so in this example, it will always print 3 characters, and I set index to 3.

The next line writes the string abc into sendbuf, starting at position 3, and the line after that prints the value of i into position 6. Finally, I add the \\n for the new line, and the last line is the null terminator. I don't put index++ because I don't care about index any more, so don't need to bother incrementing it. I put \\0 (as did you above) instead of 0, simply because \\0 reminds the programmer that we're dealing with a char array, not an integer array.

Again, all these lines aren't necessary, but since some of the issue seemed to be understanding what printf does, I thought it would be helpful.

Answer 2

这可以工作：

snprintf(sendbuf, sizeof(sendbuf), "%.1fabc%d", count, i);

Answer 3

Add a byte to the output array. The first element needs 3 characters (bytes) not one.

sprintf(sendBuffer,"%0.1f",count);
sendBuffer[3] = 'a';
....