这取决于您所说的性能...就操作数量而言,您的第一个示例是最好的(操作数量最少)...您的第二个和第三个示例几乎完全相同。
echo "Hello" . "World! <br/>";HERE 的操作码
Finding entry points
Branch analysis from position: 0
Return found
filename: /in/oYvSm
function name: (null)
number of ops: 3
compiled vars: none
line # * op fetch ext return operands
---------------------------------------------------------------------------------
3 0 > CONCAT ~0 'Hello', 'World%21+%3Cbr%2F%3E'
1 ECHO ~0
5 2 > RETURN 1
branch: # 0; line: 3- 5; sop: 0; eop: 2
path #1: 0,
echo "Hello"; echo "World!", "<br/>";HERE 的操作码
Finding entry points
Branch analysis from position: 0
Return found
filename: /in/nMufh
function name: (null)
number of ops: 4
compiled vars: none
line # * op fetch ext return operands
---------------------------------------------------------------------------------
3 0 > ECHO 'Hello'
1 ECHO 'World%21'
2 ECHO '%3Cbr%2F%3E'
5 3 > RETURN 1
branch: # 0; line: 3- 5; sop: 0; eop: 3
path #1: 0,
echo "Hello", "World!", "<br/>";HERE 的操作码
Finding entry points
Branch analysis from position: 0
Return found
filename: /in/LnPaY
function name: (null)
number of ops: 4
compiled vars: none
line # * op fetch ext return operands
---------------------------------------------------------------------------------
3 0 > ECHO 'Hello'
1 ECHO 'World%21'
2 ECHO '%3Cbr%2F%3E'
4 3 > RETURN 1
branch: # 0; line: 3- 4; sop: 0; eop: 3
path #1: 0,
因此,您可以清楚地看到您在第二个和第三个示例中使用了一个额外的echo 操作。但在这些示例中,性能(读取速度)几乎可以忽略不计。