char字符串中的ascii代码

Question

我如何从datagenerator datapoint*25+65获得单个字符，如a，但我想在str = abcdefg （无论哪个字母）？ datapoint创建一个介于0.0和1.0之间的值。 单个字符的程序：

char str;
for(int n; n<60; n++)
{
    str=datapoint*25+65;
    str++;
}
str = '/0';

问题我不知道如何使用这个设置获得像abcd这样的字符串，而不仅仅是像str中的单个字母。

Answer 1

尝试这个：

char str[60 + 1]; /* Make target LARGE enough. */
char * p = str; /* Get at pointer to the target's 1st element. */
for(int n = 0; /* INITIALISE counter. */ 
    n<60; 
    n++)
{
  *p = datapoint*25+65; /* Store value by DE-referencing the pointer before
                         assigning the value to where it points. */
  p++; /* Increment pointer to point to next element in target. */
}
*p = '\0'; /* Apply `0`-terminator using octal notation, 
              mind the angle of the slash! */

puts(str); /* Print the result to the console, note that it might (partly) be
              unprintable, depending on the value of datapoint. */

没有指向当前元素但使用索引的替代方法：

char str[60 + 1]; /* Make target LARGE enough. */
for(int n = 0; /* INITIALISE counter. */ 
    n<60; 
    n++)
{
  str[n] = datapoint*25+65; /* Store value to the n-th element. */
}
str[n] = '\0'; /* Apply `0`-terminator using octal notation, 
                  mind the angle of the slash! */

Answer 2

char str; 
/* should be char str[61] if you wish to have 60 chars
 * alternatively you can have char *str
 * then do str=malloc(61*sizeof(*str));
 */

for(int n; n<60; n++)
/* n not initialized -> should be initialized to 0,
 * I guess you wish to have 60 chars
 */
{
    *(str+n)=datapoint*25+65; 
/* alternative you can have str[n]=datapoint*25+65;
 * remember datapoint is float
 * the max value of data*25+65 is 90 which is the ASCII correspondent for 
 * letter 'Z' ie when datapoint is 1.0

 * It is upto you how you randomize datapoint.
 */
}
str[60] = '\0'; // Null terminating the string.
/* if you have used malloc
 * you may do free(str) at the end of main()
 */

Answer 3

您必须了解，因为您使用的是chars，所以一次只能存储一个字符。 如果要存储多个字符，请使用字符串类或c字符串（字符数组）。 另外，请确保初始化str和n的值。 例如：

str = 'a';
n = 0;