target.def (fusion_priority): Wrap code with @smallexample.

	* target.def (fusion_priority): Wrap code with @smallexample.
	* doc/tm.texi: Regenerated.

From-SVN: r218301

diff --git a/gcc/ChangeLog b/gcc/ChangeLog
index ceb9e108a9f58be5a6ebdf08fb600ce938c1cc5e..cfc6628797931331b119c33ad3fa48b065785b24 100644 (file)
--- a/gcc/ChangeLog
+++ b/gcc/ChangeLog
@@ -1,3 +1,8 @@
+2014-12-03  Bin Cheng  <bin.cheng@arm.com>
+
+       * target.def (fusion_priority): Wrap code with @smallexample.
+       * doc/tm.texi: Regenerated.
+
 2014-12-03  Manuel López-Ibáñez  <manu@gcc.gnu.org>
 
        * diagnostic.c (diagnostic_show_locus): Honor override_column when

diff --git a/gcc/doc/tm.texi b/gcc/doc/tm.texi
index c54fc71788c96bc4a303b7dffdd8a956835881ed..b9a7251d6a1033347a2cebdc0d456a2f47ab341b 100644 (file)
--- a/gcc/doc/tm.texi
+++ b/gcc/doc/tm.texi
@@ -6797,6 +6797,7 @@ instructions.
 
 Given below example:
 
+@smallexample
     ldr r10, [r1, 4]
     add r4, r4, r10
     ldr r15, [r2, 8]
@@ -6805,6 +6806,7 @@ Given below example:
     add r4, r4, r11
     ldr r16, [r2, 12]
     sub r5, r5, r16
+@end smallexample
 
 On targets like ARM/AArch64, the two pairs of consecutive loads should be
 merged.  Since peephole2 pass can't help in this case unless consecutive
@@ -6812,19 +6814,22 @@ loads are actually next to each other in instruction flow.  That's where
 this scheduling fusion pass works.  This hook calculates priority for each
 instruction based on its fustion type, like:
 
-    ldr r10, [r1, 4]  ; fusion_pri=99,  pri=96   
-    add r4, r4, r10   ; fusion_pri=100, pri=100  
-    ldr r15, [r2, 8]  ; fusion_pri=98,  pri=92   
-    sub r5, r5, r15   ; fusion_pri=100, pri=100  
-    ldr r11, [r1, 0]  ; fusion_pri=99,  pri=100  
-    add r4, r4, r11   ; fusion_pri=100, pri=100  
-    ldr r16, [r2, 12] ; fusion_pri=98,  pri=88   
-    sub r5, r5, r16   ; fusion_pri=100, pri=100  
+@smallexample
+    ldr r10, [r1, 4]  ; fusion_pri=99,  pri=96
+    add r4, r4, r10   ; fusion_pri=100, pri=100
+    ldr r15, [r2, 8]  ; fusion_pri=98,  pri=92
+    sub r5, r5, r15   ; fusion_pri=100, pri=100
+    ldr r11, [r1, 0]  ; fusion_pri=99,  pri=100
+    add r4, r4, r11   ; fusion_pri=100, pri=100
+    ldr r16, [r2, 12] ; fusion_pri=98,  pri=88
+    sub r5, r5, r16   ; fusion_pri=100, pri=100
+@end smallexample
 
 Scheduling fusion pass then sorts all ready to issue instructions according
 to the priorities.  As a result, instructions of same fusion type will be
 pushed together in instruction flow, like:
 
+@smallexample
     ldr r11, [r1, 0]
     ldr r10, [r1, 4]
     ldr r15, [r2, 8]
@@ -6833,6 +6838,7 @@ pushed together in instruction flow, like:
     sub r5, r5, r15
     add r4, r4, r11
     sub r5, r5, r16
+@end smallexample
 
 Now peephole2 pass can simply merge the two pairs of loads.
 

diff --git a/gcc/target.def b/gcc/target.def
index 7c0296d58e47a9825d7cdfcd1bdcbb2be4c920dc..647ebbe7bb6093c4cf568d80bcc3acb21332ded5 100644 (file)
--- a/gcc/target.def
+++ b/gcc/target.def
@@ -1555,6 +1555,7 @@ instructions.\n\
 \n\
 Given below example:\n\
 \n\
+@smallexample\n\
     ldr r10, [r1, 4]\n\
     add r4, r4, r10\n\
     ldr r15, [r2, 8]\n\
@@ -1563,6 +1564,7 @@ Given below example:\n\
     add r4, r4, r11\n\
     ldr r16, [r2, 12]\n\
     sub r5, r5, r16\n\
+@end smallexample\n\
 \n\
 On targets like ARM/AArch64, the two pairs of consecutive loads should be\n\
 merged.  Since peephole2 pass can't help in this case unless consecutive\n\
@@ -1570,19 +1572,22 @@ loads are actually next to each other in instruction flow.  That's where\n\
 this scheduling fusion pass works.  This hook calculates priority for each\n\
 instruction based on its fustion type, like:\n\
 \n\
-    ldr r10, [r1, 4]  ; fusion_pri=99,  pri=96   \n\
-    add r4, r4, r10   ; fusion_pri=100, pri=100  \n\
-    ldr r15, [r2, 8]  ; fusion_pri=98,  pri=92   \n\
-    sub r5, r5, r15   ; fusion_pri=100, pri=100  \n\
-    ldr r11, [r1, 0]  ; fusion_pri=99,  pri=100  \n\
-    add r4, r4, r11   ; fusion_pri=100, pri=100  \n\
-    ldr r16, [r2, 12] ; fusion_pri=98,  pri=88   \n\
-    sub r5, r5, r16   ; fusion_pri=100, pri=100  \n\
+@smallexample\n\
+    ldr r10, [r1, 4]  ; fusion_pri=99,  pri=96\n\
+    add r4, r4, r10   ; fusion_pri=100, pri=100\n\
+    ldr r15, [r2, 8]  ; fusion_pri=98,  pri=92\n\
+    sub r5, r5, r15   ; fusion_pri=100, pri=100\n\
+    ldr r11, [r1, 0]  ; fusion_pri=99,  pri=100\n\
+    add r4, r4, r11   ; fusion_pri=100, pri=100\n\
+    ldr r16, [r2, 12] ; fusion_pri=98,  pri=88\n\
+    sub r5, r5, r16   ; fusion_pri=100, pri=100\n\
+@end smallexample\n\
 \n\
 Scheduling fusion pass then sorts all ready to issue instructions according\n\
 to the priorities.  As a result, instructions of same fusion type will be\n\
 pushed together in instruction flow, like:\n\
 \n\
+@smallexample\n\
     ldr r11, [r1, 0]\n\
     ldr r10, [r1, 4]\n\
     ldr r15, [r2, 8]\n\
@@ -1591,6 +1596,7 @@ pushed together in instruction flow, like:\n\
     sub r5, r5, r15\n\
     add r4, r4, r11\n\
     sub r5, r5, r16\n\
+@end smallexample\n\
 \n\
 Now peephole2 pass can simply merge the two pairs of loads.\n\
 \n\