*string = toupper(*string++)报错的原因在哪里
*string = toupper(*string++)出错的原因在哪里？
使用
使用
源文件如下：
------解决思路----------------------
看置顶帖
------解决思路----------------------
置顶贴有说明.http://bbs.****.net/topics/370153775
如果看不懂就, 就永远把自加自减单独写一行.
------解决思路----------------------
------解决思路----------------------
同意5L，在彻底掌握++之前，实践时只把++单独写一行。
使用
*string = toupper(*string++)无法执行大写转换操作。
使用
*string = toupper(*string);与之前的语句有什么区别？
string++;
源文件如下：
/*
*
*/
#include <stdio.h>
#include <string.h>
#include <ctype.h>
void toUpper(char *string);
void showString(const char *string);
int main(void)
{
char string[80] = "I love the C Programming.\n";
toUpper(string);
showString(string);
return 0;
}
void toUpper(char *string)
{
while('\0' != *string){
*string = toupper(*string);
string++;
}
}
void showString(const char *string)
{
while('\0' != *string)
putchar(*string++);
}
------解决思路----------------------
看置顶帖
------解决思路----------------------
置顶贴有说明.http://bbs.****.net/topics/370153775
如果看不懂就, 就永远把自加自减单独写一行.
------解决思路----------------------
//C++ Operators
// Operators specify an evaluation to be performed on one of the following:
// One operand (unary operator)
// Two operands (binary operator)
// Three operands (ternary operator)
// The C++ language includes all C operators and adds several new operators.
// Table 1.1 lists the operators available in Microsoft C++.
// Operators follow a strict precedence which defines the evaluation order of
//expressions containing these operators. Operators associate with either the
//expression on their left or the expression on their right; this is called
//“associativity.” Operators in the same group have equal precedence and are
//evaluated left to right in an expression unless explicitly forced by a pair of
//parentheses, ( ).
// Table 1.1 shows the precedence and associativity of C++ operators
// (from highest to lowest precedence).
//
//Table 1.1 C++ Operator Precedence and Associativity
// The highest precedence level is at the top of the table.
//+------------------+-----------------------------------------+---------------+
//
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Operator
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Name or Meaning
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Associativity
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//+------------------+-----------------------------------------+---------------+
//
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::
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Scope resolution
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None
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//
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::
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Global
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None
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//
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[ ]
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Array subscript
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Left to right
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//
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( )
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Function call
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Left to right
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//
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( )
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Conversion
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None
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//
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.
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Member selection (object)
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Left to right
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//
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->
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Member selection (pointer)
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Left to right
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//
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++
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Postfix increment
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None
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//
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--
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Postfix decrement
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None
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//
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new
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Allocate object
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None
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//
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delete
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Deallocate object
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None
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//
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delete[ ]
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Deallocate object
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None
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//
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++
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Prefix increment
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None
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//
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--
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Prefix decrement
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None
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//
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*
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Dereference
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None
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//
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&
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Address-of
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None
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//
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+
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Unary plus
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None
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//
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-
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Arithmetic negation (unary)
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None
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//
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!
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Logical NOT
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None
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//
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~
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Bitwise complement
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None
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//
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sizeof
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Size of object
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None
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//
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sizeof ( )
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Size of type
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None
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//
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typeid( )
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type name
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None
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//
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(type)
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Type cast (conversion)
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Right to left
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//
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const_cast
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Type cast (conversion)
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None
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//
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dynamic_cast
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Type cast (conversion)
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None
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//
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reinterpret_cast
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Type cast (conversion)
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None
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//
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static_cast
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Type cast (conversion)
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None
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//
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.*
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Apply pointer to class member (objects)
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Left to right
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//
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->*
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Dereference pointer to class member
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Left to right
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//
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*
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Multiplication
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Left to right
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//
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/
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Division
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Left to right
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//
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%
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Remainder (modulus)
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Left to right
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//
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+
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Addition
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Left to right
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//
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-
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Subtraction
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Left to right
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//
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<<
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Left shift
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Left to right
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//
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>>
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Right shift
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Left to right
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//
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<
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Less than
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Left to right
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//
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>
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Greater than
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Left to right
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//
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<=
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Less than or equal to
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Left to right
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//
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>=
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Greater than or equal to
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Left to right
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//
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==
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Equality
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Left to right
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//
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!=
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Inequality
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Left to right
------解决思路----------------------
//
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&
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Bitwise AND
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Left to right
------解决思路----------------------
//
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^
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Bitwise exclusive OR
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Left to right
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//
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------解决思路----------------------
------解决思路----------------------
Bitwise OR
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Left to right
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//
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&&
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Logical AND
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Left to right
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//
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------解决思路----------------------
------解决思路----------------------
Logical OR
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Left to right
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//
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e1?e2:e3
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Conditional
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Right to left
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//
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=
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Assignment
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Right to left
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//
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*=
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Multiplication assignment
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Right to left
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//
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/=
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Division assignment
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Right to left
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//
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%=
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Modulus assignment
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Right to left
------解决思路----------------------
//
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+=
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Addition assignment
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Right to left
------解决思路----------------------
//
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-=
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Subtraction assignment
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Right to left
------解决思路----------------------
//
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<<=
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Left-shift assignment
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Right to left
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//
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>>=
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Right-shift assignment
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Right to left
------解决思路----------------------
//
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&=
------解决思路----------------------
Bitwise AND assignment
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Right to left
------解决思路----------------------
//
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------解决思路----------------------
=
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Bitwise inclusive OR assignment
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Right to left
------解决思路----------------------
//
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^=
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Bitwise exclusive OR assignment
------解决思路----------------------
Right to left
------解决思路----------------------
//
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,
------解决思路----------------------
Comma
------解决思路----------------------
Left to right
------解决思路----------------------
//+------------------+-----------------------------------------+---------------+
------解决思路----------------------
同意5L，在彻底掌握++之前，实践时只把++单独写一行。